4 Integration on \(\mathbb R^n\)

This section explores measurable functions and Lebesgue integration on \(\mathbb R^n\).

  1. Measurable functions on \(\mathbb R^n\): approximation by simple functions, conditions for measurability, limit of measurable functions.
    • Axler, chapter 3.
  2. Littlewood’s three principles: Egorov’s and Luzin’s theorems.
    • Axler, chapter 3.
  3. Lebesgue integration: properties of of the Lebesgue integral, \(L^1\) norm.
    • Rudin, chapter 11.
  4. Convergence theorems: monotone and dominated convergence theorems, Fatou’s lemma, relation to Riemann integral.
    • Rudin, chapter 11.

Key takeaways:

  1. Limit theorems:
    • Every real measurable function is the pointwise limit of measurable simple functions 4.1.
    • Pointwise limit of measurable functions is measurable 4.5.
  2. Littlewood’s three principles:
    • Every measurable set on \(\mathbb R\) can be approximated up to arbitrary but finite tolerance by a finite union of open intervals.
    • Egorov’s theorem 4.3: for every finite (but arbitrary) tolerance to approximating a domain with finite measure, pointwise convergence a.e. is uniform convergence.
    • Luzin’s theorem 4.4: every \(\mathbb R\to \mathbb R\) measurable function is continuous in the subspace topology of a finite but arbitrary closed approximation to \(\mathbb R\).
  3. Convergence theorems:
    • Monotone convergence theorem: pointwise monotone limit of nonnegative measurable functions can be exchanged with integration.
    • Fatou’s lemma: \(\lim\) of integral \(\leq\) integral of \(\liminf\) for nonnegative measurable sequences.
    • Dominated convergence theorem: pointwise dominated limit can be exchanged with integration.
  4. General technique: prove for characteristic functions, then generalize using approximation and limit theorems.
  5. Fatou’s lemma can be used to2. Sufficient conditions for measurability: semicontinuous, monotonic, the pointwise limit of a measurable sequence, or equals a measurable function a.e.
  6. Properties of the Lebesgue integral:
    • \(f\) integrable iff \(|f|\) integrable.
    • disjoint additivity w.r.t. domain.
    • linear and order-preserving.
    • sets of measure zero are negligible.

Measurable functions on \(\mathbb R^n\)

Simple functions

Definition 4.1 (characteristic function) Given \(E\subset \Omega\), the characteristic function \(\chi_E:\Omega\to \mathbb R\) is \[ \chi_E(x) = \begin{cases} 1 & x\in E \\ 0 & x\notin E \end{cases} \]

Proposition 4.1 Given \((\Omega, \Sigma), E\subset \Omega\), then \(\chi_E\) is \(\Sigma\)-measurable iff \(E\in \Sigma\).

Proof: \(\chi_E^{-1}(B) = \{E, \Omega-E, \Omega, \emptyset\}\) depending on how \(B\subset \mathbb R\) intersects with \(\{0, 1\}\).

Definition 4.2 (simple function) A simple function is a finite sum below, where each \(E_k\) is a finite-measurable set. \[ f = \sum_{k=1}^N a_k \chi_{E_k} \]

Theorem 4.1 (simple-function approximation) Given a real function \(f:\Omega\to \mathbb R\), there exists a sequence \(\{s_n\}\) of simple functions such that \(s_n(x)\to f(x)\) point-wise. If \(f\) is measurable, then \(\{s_n\}\) may be chosen to be a sequence of measurable functions. If \(f\geq 0\), then \(\{s_n\}\) may be chosen to be a monotonically increasing sequence.

Proof idea: in the \(n\)-th approximation, approximate \(f\geq 0\) (general case follows by \(f=f^+-f^-\)) up to value \(n\) by (1) divide \(f^{-1}([0, n])\) into \(2^{n}\) pieces, (2) assign a dominated to each of \(2^n\) intervals, and (3) approximate \(f(x)>n\) with \(n\).
Proof

Given \(f\geq 0\), define \(F_n = f^{-1}([n, \infty))\) and \[ E_{n, i} = f^{-1} \left(\left[2^{-n}(i-1), 2^{-n}i \right]\right) \] For \(n=1, 2, \cdots\) and \(i=1, 2, \cdots, n2^n\), choose \[ s_n = n\chi_{F_n} + \sum_{i=1}^{n2^n} \dfrac{i-1}{2^n} \chi_{E_{n, i}} \] Note that \(E_{n, i}\) is the pre-image of the \(i\)-th slice of nonnegative real blocks of size \(2^{-n}\), and the first term assigns \(\inf f=2^{-n}(i-1)\) on the corresponding intervals. The second term just assigns approximates with \(n\) all places where \(f(x)\geq n\). The second term is reserved for cases where \(f(x)=\infty\). If \(f\) is measurable, the domains of the characteristic decomposition are measurable so \(s_n\) is measurable. The general case follows by \(f=f^+-f^-\).

Conditions for measurability

Example 4.1 If \(\Sigma=\{\emptyset, \Omega\}\), then the only \(\Sigma\)-measurable functions from \(\Omega\to \mathbb R^n\) are constant functions. On the other extreme, if \(\Sigma=2^\Omega\), then every function is measurable. For a nontrivial example, if \(\Sigma=\{\emptyset, (-\infty, 0), [0, \infty), \mathbb R\}\) then \(f:\mathbb R\to \mathbb R\) is measurable iff it is constant on \((-\infty, 0), [0, \infty)\).

Proposition 4.2 (semi-continuity implies measurability) Every (lower or upper) semi-continuous \(R^n\)-valued function (definition 1.4) is measurable.

Proof: Recall proposition 2.6. The Borel \(\sigma\)-algebra is generated by half-rectangles \((\infty, t\in \mathbb R^n)\) or \((t, \infty)\)

The explicit proof below demonstrates the relation between continuity and measurability.

Proposition 4.3 (continuity implies Borel measurability) Every continuous function \(f:\Omega\to \mathbb R^n\) defined on a Borel subset \(\Omega\subset \mathbb R^m\) is a Borel-measurable function.

Proof Consider the preimage of \(x>a\) (\(x, a\in \mathbb R^n\) and inequality holds elementwise). For any such \(a\), consider \[ f^{-1}(x>a) = \left[\bigcup_{x\in f^{-1}(x>a)} (x-\delta_x, x+\delta_x) \right]\cap \Omega \] Given every \(x\in \mathbb R^m: f(x)>a\), by continuity \(\exists \delta_x\in \mathbb R\) such that \(f((x-\delta_x, x+\delta_x))>a\). The term in the paranthesis is an open set, hence its intersection with the Borel set \(\Omega\) is Borel.

\(f:(\Omega\subset \mathbb R)\to \mathbb R\) is increasing if \(x < y\implies f(x)\leq f(y)\); it is monotonic if \(f\) or \(-f\) is increasing.

Proposition 4.4 if \(f:\Omega\to \mathbb R\) is monotonic and \(\Omega\) is Borel, then \(f\) is measurable.

Proof Fix \(a\in \mathbb R\), then for \(b=\inf f^{-1}((a, \infty))\), monotonicity implies \[ f^{-1}((a, \infty)) = (b, \infty)\cap \Omega \text{ or } [b, \infty)\cap \Omega \]

Proposition 4.5 (preservation of measurability under limits) Suppose a measurable sequence \(\{f_j\}\to f:(\Omega, \Sigma)\to \mathbb R^n\) pointwise is \(\Sigma\)-measurable, then \(f\) is \(\Sigma\)-measurable.

Proof Unrolling the definition of the limit: for any \(x\in \Omega, a\in \mathbb R^n\), \(f(x)>a\) iff \[\begin{align} & \lim_{k\to \infty} f_k(x) > a \\ \iff & \exists j>0, \exists N: f_{\forall k>N}(x)>a+1/j \\ \iff & x\in \bigcup_{j=1}^\infty \bigcup_{N=1}^\infty \bigcap_{k=1}^\infty f_k^{-1}(y>a+1/j) \end{align}\] Note that we only need some \(j\) to exist. Each \(\exists\) translates to \(\bigcup\), and each \(\forall\) translates to \(\bigcap\). Then the preimage of \(y>a\) the measurable set above.

Definition 4.3 (extended Borel) \(A\subset [-\infty \infty]\) is a Borel set if \(A\cap \mathbb R\) is Borel. Extended measurability is defined similarly.

Theorem 4.2 (variational limits are measurable) Given measurable \(\{f_k: (\Omega, \Sigma)\to \mathbb R\}\), the following functions are also measurable: \[ g(x) = \inf_k f_k(x), \quad h(x) = \sup_k f_k(x) \]

Proof It suffices to verify for the \(\sup\) case that \[\begin{align} h(x) > a &\implies \exists k: f_k(x) > a \\ h^{-1}((a, \infty]) &= \bigcup_{k=1}^\infty f_k^{-1}((a, \infty]) \end{align}\]

Proposition 4.6 (a.e. measurability) Given \(f\) measurable and \(f=g\) a.e. (they differ on a set of measure \(0\) on the domain), then \(g\) is measurable.

Proof The symmetric difference of the pre-images \(f^{-1}, g^{-1}\) is dominated by a set of measure \(0\), then by \(f\)-measurability \(g^{-1}(A)\) is measurable for \(A\) measurable (proposition 3.7).

Littlewood’s three principles

Littlewood’s three principles (at least the ones we prove) apply to \(\mathbb R\) and real-valued functions on the real domain.

  1. Every finite-measurable set is almost a finite union of intervals.
  2. Every measurable function is almost continuous.
  3. Every pointwise convergent sequence of measurable functions is almost uniformly convergent.

It is convenient to first introduce some notation for monotone limits.

Definition 4.4 (sequence of sets) An increasing sequence of sets \(\{A_n\}\) is a sequence \[ A_1\subset A_2\subset \cdots \to A = \bigcup_{n=1}^\infty A_n \] also written \(A_j\nearrow A\). Similarly, a decreasing sequence of sets is written \(A_j\searrow A\). \[ A_1\supset A_2\supset \cdots \to A = \bigcap_{n=1}^\infty A_n \]

Proposition 4.7 (Littlewood's first principle) Given measurable \(E\subset \mathbb R\) with \(\mu(E)<\infty\), there exists a finite union \(O\) of intervals such that \(\mu(O\Delta E) < \epsilon\) for every \(\epsilon\).

Proof idea: prove for open \(E\) and apply theorem 3.3.
Proof Every open set \(O\) is the interval of countably many disjoint open intervals (this is because every disjoint interval contains a rational number, of which there are countably many), then \(\mu(O) = \sum_{j=1}^n \mu((a_j, b_j))\). Thus for every \(\epsilon>0\) we can choose a \(\epsilon\)-approximation with finite intervals. Apply theorem 3.3 to obtain the result for general measurable sets.

To be precise, the third principle is formulated as follows:

Theorem 4.3 (Egorov's theorem) Consider a measurable sequence \(f_k\to f\) a.e. on a finite-measurable set \(E\) with \(\mu(E)<\infty\). For every \(\epsilon>0\), there exists a closed set \(A_\epsilon \subset E\) such that \(f_k\to f\) uniformly and \(\mu(E-A_\epsilon)<\epsilon\) on \(A_\epsilon\).

Proof idea: given domain tolerance \(\epsilon\), construct by pointwise convergence a family \(F_n\) of \((\epsilon 2^{-n})\)-domain approximations on which \(f_j\) will uniformly \(1/n\)-approximate \(f\) on \(F_n\). Then \(f_j\to f\) uniformly on the intersection \(A_\epsilon\) of \(F_n\), which is also an \(\epsilon\)-domain approximation.
Proof

Note that we can’t simply use compactness here since the domain approximation may be unbounded.

  1. Translate pointwise convergence: fixing \(n\in \mathbb N\), for every \(x\in X\) there exists \(m\in \mathbb N\) such that \(f_{\forall k\geq m}\) will \(\epsilon\)-approximate \(f\) at \(x\), then \[ X = \bigcup_{m=1}^\infty A_{m, n}, \quad A_{m, n} = \bigcap_{k=m}^\infty \{x: |f_k(x) - f(x)| < 1/n\} \] Here \(A_{m, n}\) is the domain on which \(f_{\forall k\geq m}\) will \(1/n\)-approximate \(f\).
  2. Given domain tolerance \(\epsilon>0\), for each \(n>0\) we can find \(m_n\) such that \[ \mu(X) - \mu(A_{m_n, n}) < \dfrac\epsilon {2^n} \] This definition implies that \(f_{\forall j\geq m_n}\) will \(1/n\)-approximate \(f\) on \(A_{m_n, n}\).
  3. Let \(A_\epsilon = \bigcap_{n=1}^\infty A_{m_n, n}\); then for every \(\epsilon'>1/N>0\), \(f\) will be \(\epsilon'\)-approximated by \(f_{j\geq m_N}\) on \(A_\epsilon \subset A_{M_N, N} \subset A_\epsilon\).
  4. Argue for \(\epsilon\)-approximation of the domain: \[\begin{align} \mu(\Omega - A_\epsilon) &= \mu \bigcup_{n=1}^\infty \Omega - A_{m_n, n} \leq \sum_{n=1}^\infty \epsilon/2^n < \epsilon \end{align}\]
  5. We can choose \(A_\epsilon\) to be closed by the characterization of measurable sets (theorem 3.3).

Similarly, the precise statement of the second principle is as follows. Note that does not claim that \(f\) is pointwise continuous on \(F_\epsilon\) in the topology of \(\mathbb R\), only that \(f_{|F_\epsilon}\) is continuous in the subspace topology.

Lemma 4.1 Given a Borel-measurable \(f:\mathbb R\to \mathbb R\), for every \(\epsilon>0\) there exists a closed \(F\subset \mathbb R\) with \(|\mathbb R- F|<\epsilon\) such that \(f_{|F}\) is continuous on \(F\).

Proof idea: establish for (1) simple functions, (2) arbitrary functions on compact interval, using pointwise approximation by simple functions and Egorov’s theorem, and (3) extended domain by pasting local chunks. a.e.
Proof

First assume \(f = \sum_{j=1}^N \alpha_j \chi_{E_j}\). By theorem 3.3 choose \(U_j\supset E_j\supset F_j\) with \(U_j\) open, \(F_j\) closed, and both are \((\epsilon/2n)\)-approximations of \(E_j\). The closed set is the union of each \(F_j\) and \(U_j^c\): \[ E = F \cup (\mathbb R- U), \quad F = \bigcup_{j=1}^N F_j, \quad U = \bigcup_{j=1}^N U_j \] both \(F, U^c\) are closed and \(\mu(\mathbb R-E) < \epsilon\). Continuity on \(E\) follows by \(f\) being constantly zero on disjoint closed intervals, so the pasting lemma applies.

Next consider general \(f\) on a compact interval. Every \(f\) is the pointwise limit of simple functions (theorem 4.1) and their uniform limit on a closed subset (Egorov’s theorem 4.3) thus continuous.

To extend the domain, construct \(\epsilon\cdot 2^{-n-1}\)-approximations to \(f\) on \([-n, 1-n]\) and \([n-1, n]\) and apply the pasting lemma.

Theorem 4.4 (Luzin's theorem) Given a Borel measurable \(f:(E\subset \mathbb R)\to \mathbb R\), for every \(\epsilon>0\) there exists a closed \(\epsilon\)-approximation \(F \subset E\) such that \(f_{|F}\) is continuous in the subspace topology.

Proof: Apply lemma 4.1 to the continuous extension of \(f\) to all of \(\mathbb R\).

Lebesgue integration

Definition 4.5 (function support) The support of a measurable function \(f\) is \[ \mathrm{supp}(f) = \overline{f^{-1}(\{0\}^C)} \] the closure of domain on which \(f\) does not vanish.

Definition 4.6 (S-partition) Given a \(\sigma\)-algebra \(\Sigma\) on \(\Omega\), a \(S\)-partition of \(\Omega\) is a finite collection \(A_1, \cdots, A_m\in \Sigma\) of disjoint sets such that \(\bigsqcup A_j = \Omega\).

Definition 4.7 (lower Lebesgue sum) Given measure space \((\Omega, \Sigma, \mu)\), an \(\Sigma\)-measurable function \(f:\Omega\to [0, \infty]\), and a \(\Sigma\)-partition \(P=\{A_1, \cdots, A_m\}\) of \(\Omega\), the lower Lebesgue sum is \[ \mathcal L_\mu(f, P) = \sum_{j=1}^m \mu(A_j)\inf_{A_j} f \]

Definition 4.8 (nonnegative integral) The integral of a non-negative function \(f:\Omega\to [0, \infty]\) w.r.t. the measure \((\Sigma, \mu)\) is the following \(\sup\) with \(P\) over all S-partitions of \(\Omega\): \[ \int f\, d\mu = \sup_P \mathcal L_\mu(f, P) \]

Proposition 4.8 (base case for Lebesgue integral) Given \((\Omega, \Sigma, \mu)\) and measurable \(E\in \Sigma\), we have \(\int \chi_E\, d\mu = \mu(E)\).

Proof We have \(\mathcal L_\mu(\chi_E, \{E, \Omega-E\}) = \mu(E)\), then \(\int f\, d\mu \geq \mu(E)\). Conversely, given any S-partition \(P=\{A_1, \cdots, A_m\}\) of \(\Omega\), for each \(j\) \[ \mu(A_j) \inf_{A_j} \chi_E = \begin{cases} \mu(A_j) & A_j\subset E \\ 0 & \text{otherwise} \end{cases} \] then \(\mathcal L(\chi_E, P) = \sum_{A_j\subset E} \mu(A_j) \leq \mu(E)\).

Proposition 4.9 (integral of a simple function) Given a measure space \((\Omega, \Sigma, \mu)\) and disjoint \(E_1, \cdots E_n\), we have \[ \int \left(\sum_{k=1}^n c_k \chi_{E_k}\right)\, d\mu = \sum_{k=1}^n c_k \mu(E_k) \]

Proof The LHS integral is lower-bounded by the \(\Sigma\)-partition \(\{E_j\}\cup \{\Omega-\bigcup E_j\}\). The converse follows similar to the last proposition.

Corollary 4.1 (equivalent definition of nonnegative integral) Given \((\Omega, \Sigma, \mu)\) and measurable \(f:\Omega\to [0, \infty]\), the integral definition 4.8 is equivalent to \[ \int f\, d\mu = \sup_P \mathcal L_\mu(f, P) = \sup_{s\leq f} \int s\, d\mu \] where the last \(\sup\) is taken over all simple function \(f\) dominated by \(f\): \(\forall x\in \Omega: f(x)\geq s(x)\).

Proof: Follows from \(\mathcal L_\mu(f, P)\) being the integral of a simple function.

Definition 4.9 (Lebesgue integral) Given \((\Omega, \Sigma, \mu)\) and measurable \(f(x)\in [-\infty, \infty]\) with \(f=f^+-f^-\) and \(f^+, f^-\geq 0\). If at least one of \(\int f^+\, d\mu, \int f^-\, d\mu\) is finite, then the Lebesgue integral of \(f\) is \[ \int f\, d\mu = \int f^+\, d\mu - \int f^-\, d\mu \]

Note that, for example, the integral of the following function is not defined: \[ f(x) = \begin{cases} 1 & x\geq 0 \\ -1 & x < 0 \end{cases} \]

Definition 4.10 (integral metric, integrable space) Fix \((\Omega, \Sigma, \mu)\); if \(f:\Omega\to [-\infty, \infty]\) is measurable, then the \(L^1\)-norm of \(f\) is \[ \|f_1\| = \int |f|\, d\mu \] The Lebesgue space \(L^1(\mu)\) is also defined as \(L^1(\mu) = \{f:\Omega\to \mathbb R\text{ satisfying } \|f\|_1 < \infty \}\).

Theorem 4.5 (properties of integration) The following properties are evident for \(f, g\in L^1(\mu)\):

  1. Linearity: \(\int cf+g\, d\mu = c\int f\, d\mu + \int g\, d\mu\).
  2. Order-preserving: \(f\leq g\implies \int f\, d\mu \leq \int g\, d\mu\).
  3. \(\left|\int f\, d\mu\right| \leq \int |f|\, d\mu\).
Proof General idea is to establish for simple functions, then show that the proof commutes with integral construction. For (2), use \(\int g\, d\mu - \int f\, d\mu = \int (g-f)\, d\mu\). For (3), \[\begin{align} \left|\int f\, d\mu\right| &= \left| \int f^+\, d\mu - \int f^-\, d\mu\right| \leq \int f^+\, d\mu + \int f^-\, d\mu \\ &= \int (f^++f^-)\, d\mu = \int |f|\, d\mu \end{align}\]

Definition 4.11 (integration on subset) Given \((\Omega, \Sigma, \mu)\) and \(E\in \Sigma\) and measurable \(f:\Omega\to [-\infty, \infty]\), the subset integral is \[ \int_E f\, d\mu = \int \chi_E f\, d\mu \]

Proposition 4.10 (integral bound) Given \(\int_E f\, d\mu < \infty\), \[ \left|\int_E f\, d\mu \right| \leq \mu(E) \sup_E |f| \]

Proof Let \(c=\sup_E |f|\), then \(\displaystyle \left|\int_E f\, d\mu\right| \leq \int \chi_E |f|\, d\mu \leq \int c\chi_E \, d\mu = c\mu(E)\).

Theorem 4.6 (disjoint additivity w.r.t domain) Given measurable \(f\), define the function \(\phi:\Sigma\to [-\infty, \infty]\) by \[ \phi(A) = \int_A f\, d\mu \] then for every \(A, B\in \Sigma\), we have \(\phi(A\sqcup B) = \phi(A)+\phi(B)\).

Proof Without loss of generality consider nonnegative \(f\). If \(f\) is a characteristic function, then countable additivity is equivalent to the countable subadditivity of \(\mu\). The general proof just demonstrates the commutativity of countable subadditivity with the \(\sup\) integral construction.

Corollary 4.2 (sets of zero measure are negligible in integration) Given \(A, B\in \Sigma\) such that \(\mu(A-B)=0\), then \[ \int_A f\, d\mu = \int_B f\, d\mu \]

Proof Follows from additivity theorem 4.6 and applying \(\mu(A-B)=0\) to proposition 4.10.

The integrability of \(f\) implies that of \(|f|\), and the Lebesgue integral is often called an absolutely convergent integral.

Proposition 4.11 (several conditions for integrability)

  1. \(f\in L^1(\mu)\implies |f|\in L^1(\mu)\).
  2. \(f\) is measur_Proof:_able and \(|f|\leq g \in L^1(\mu)\implies f\in L^1(\mu)\).

Proof: for (1), recall that \(f\in L^1(\mu)\) if both \(f^+, f^-\) have finite integrals, then \(|f|=f^++f^-\) is integrable. Similarly, (2) follows from \(f^+, f^-\leq g\).

Convergence theorems

The following convergence theorems are the cornerstone of Lebesgue theory:

Theorem 4.7 (Lebesgue's monotone convergence theorem) Given \(E\) measurable, let \(\{f_n\}\) be a sequence of measurable functions such that \(0\leq f_1\leq \cdots\) on \(E\) and \(f_n\to f\), then \[ \lim_{n\to \infty} \int_E f_n \, d\mu = \int_E \lim_{n\to \infty} f_n\, d\mu \] Pointwise monotone convergence of measurable functions on a measurable set implies convergence of integral.

Proof idea: one direction of the inequality is trivial; the other is established by first proving for a dominated simple approximations of \(f\) then taking the \(\sup\).
Proof In one direction, the implied equation holds for each \(m\) thus in limit \[ f_n \leq f\implies \lim_{m\to \infty} \int f_m\, d\mu \leq \int_E \lim_{n\to \infty} f_n \, d\mu \] To establish the other direction, fix \(c\in (0, 1)\) and a dominated simple approximation \(s\leq f\). Define \[ E_n = \{x| f_n \geq c s\} \] Note that \(E_j \nearrow E\) (recall notation 4.4). For every \(n\) \[ \int_E f_n \, d\mu \geq \int_{E_n} f_n\, d\mu \geq c\int_{E_n} s\, d\mu \] Let \(n\to \infty\) and take the limit \(c\to 1\) to obtain \[ \int \lim_{n\to \infty} f_n\, d\mu \geq \lim_{n\to \infty} \int_{E_n} f_n \, d\mu \geq c\int_E s\, d\mu \] Take the \(\sup\) over all \(s\) and the simple function approximation theorem 4.1 to obtain \(F\geq \int_E f\, d\mu\).

Applying the monotone convergence theorem above to partial sums yield

Corollary 4.3 Given measurable \(E\in \Sigma\) and nonnegative measurable sequence \((f_j)\); define their sum \[ f = \sum_{n=1}^\infty f_n \] then \(\int_E f = \sum_{n=1}^\infty \int_E f_n\, d\mu\).

Lemma 4.2 (Fatou's lemma) For measurable \(E\), given nonnegative measurable sequence \((f_j)\) and \[ f(x) = \liminf_{n\to \infty} f_n(x) \] then the integral is similarly bounded \[ \int_E f\, d\mu \leq \liminf_{n\to \infty} \int_E f_n\, d\mu \]

Apply the dominated convergence theorem (carefully! expand to see details) to \(g_j = \inf_{k\geq j} f_j\).
Proof Define \(g_n = \inf_{i\geq n} f_i\), then \(g_n\) is measurable and \[ i\leq j\implies g_i\leq g_j, \quad g_n\to f \] Applying the monotone convergence theorem yields \[ \int_E g_n\, d\mu \to \int_E f\, d\mu \] Note that this does not imply equality because the monotone convergence theorem above yields the first equality, and the second inequality may be strict \[ \int_E f\, d\mu = \lim_{n\to \infty} \int_E \inf_{k\geq n} f_k\, d\mu \leq \int_E f\, d\mu = \lim_{n\to \infty} \inf_{k\geq n} \int_E f_k\, d\mu \] There exist examples (see below) where measure is lost through \(\inf f_k\) but not through the \(\inf\) of the integral.

Example 4.2 (strict inequality in Fatou's theorem) Let \(g(x) = (0\leq x\leq 1/2)^?0:1\) and \[ f_{2k}(x) = g(x), \quad f_{2k+1} = g(1-x) \] then \(\liminf_{n\to \infty} f_n = 0\) but \(\forall n, \int_0^1 f_n(x)\, dx = 1/2\).

Theorem 4.8 (Lebesgue's dominated convergence theorem) Given measurable sequence \((f_j)\to f\) pointwise and \(|f_j| \leq g\) with \(g\) measurable, then \[ \lim_{n\to \infty} \int_E f_n\, d\mu = \int_E f\, d\mu \] As a corollary, \(L^1\) is complete under dominated pointwise convergence.

Apply Fatou’s theorem to \(f_n+g\geq 0\) and \(g-f_n\geq 0\).
Proof First note that \(f_j, f\) are integrable since they’re dominated by \(g\). By \(f_n+g\geq 0\), invoke Fatou’s theorem \[ \int_E (f + g)\, d\mu \leq \liminf_{n\to \infty} \int_E (f_n+g)\, d\mu \implies \int_E f\, d\mu \leq \liminf_{n\to \infty} \int_E f_n\, d\mu \] and pull \(g\) out of both sides. Again, by \(g-f_n\geq 0\) see \[\begin{align} \int_E (g - f)\, d\mu &\leq \liminf_{n\to \infty} \int_E (g - f_n)\, d\mu \\ -\int_E f\, d\mu &\leq \liminf_{n\to \infty} \int_E -f_n\, d\mu =-\limsup_{n\to \infty} \int_E f_n\, d\mu \end{align}\] Combine the two results to obtain equality \[ \limsup_{n\to \infty} \int_E f_n\, d\mu \leq \int_E f\, d\mu \leq \liminf_{n\to \infty} \int_E f_n\, d\mu \]

Corollary 4.4 (bounded convergence theorem) Given finite-measurable \(\mu(E) < \infty\) and uniformly bounded convergent sequence \((f_n)\to f\) (i.e. \(|f_n(x)|\leq M\) for all \(n, x\)), the dominated convergence theorem applies.

Proof: Given finite measure \(\mu(E)<\infty\), the constant-\(M\) function is integrable and dominates \((f_n)\).

Comparison with Riemann integral

Denote the Riemann integral by \(\mathcal R\int_a^b f\, dx\).

Theorem 4.9 (compatibility with Riemann integral) A function \(f\) Riemann integrable on \([a, b]\) implies \(f\in L^1([a, b])\) and \[ \int_a^b f\, dx = \mathcal R \int_a^b f\, dx \]

By existence of the Riemann integral, there exists a countable sequence of iteratively refined partitions \(P_k\) such that \[ \lim_{k\to \infty} L(P_k, f) = \mathcal R \underline{\int} f\, dx, \quad \lim_{k\to \infty} U(P_k, f) = \mathcal R \bar \int f\, dx \] For each partition \(P_k\), define functions \(U_k(x\in [x_{j-1}, x_j])\) and \(L_k\) to be the sup and inf along the interval, respectively, then in the Lebesgue sense, \[ L(P_k, f) = \int L_k\, dx, \quad U(P_k, f) = \int U_k\, dx \] the functions are also monotone \(L_1 \leq L_2 \leq \cdots \leq f \leq \cdots \leq U_1\). Let \(L_k\to L, U_k\to U\), then \(L, U\) are bounded measurable functions with \(L\leq f\leq U\). Then \[ \int L\, dx = \mathcal R \underline \int f\, dx, \quad \int U\, dx = \mathcal R \bar \int f\, dx \] by the monotone convergence theorem. Further note that \(f\) is Riemann-integrable iff \(\int L\, dx = \int U\, dx\). This implies \(L=U\) a.e. since \(L\leq U\), then \(L=f=U\) a.e. and \(f\) is measurable; equality with Lebesgue integral follows.

Theorem 4.10 (Riemann integrability theorem) Given \(f\) bounded on \([a, b]\), then the Riemann integral \(\mathcal R\int_a^b f\, dx\) exists iff \(f\) is continuous a.e. on \([a, b]\).

Proof: Extending the last proof: for any \(x\notin P_k\) for any \(k\), we have \(U(x)=L(x)\) iff \(f\) is continuous at \(x\) (think about this! equivalent to saying that \(\sup\approx \inf\) arbitrarily close intervals about \(x\)); the union \(\bigcup P_k\) is countable of measure \(0\), then \(f\) is continuous a.e. iff \(L=U\) a.e. iff \(f\) is Riemann-integrable on \([a, b]\).