Functions with infinite values
Definition (Integral of a measurable function with values in \([0,\infty]\))
Let \(f : E \to [0,\infty]\) be measurable with \(E\) measurable and \(f \geq 0\). Take
\[E_{1} = \{ x \in E :\ f(x) = \infty \}.\]We define
\[\begin{aligned} R(f,E) &= \{ (x,y) \in \mathbb{R}^{d+1} :\ x \in E,\ 0 \leq y \leq f(x) \} \\ &= \{ (x,y) \in \mathbb{R}^{d+1} :\ x \in E \setminus E_{1},\ 0 \leq y \leq f(x) \} \cup \big( E_{1} \times [0,\infty) \big), \end{aligned}\]and
\[\Gamma(f,E) = \{ (x,f(x)) \in \mathbb{R}^{d+1} :\ x \in E \setminus E_{1} \}.\]Just as was done for real-valued functions, we define
\[\int_{E} f \, dx = m\big( R(f,E) \big).\]Note (The region is still measurable for infinite values)
The theory developed so far guarantees that \(R(f,E)\) is measurable whenever \(f\) and \(E\) are measurable: the first part of the decomposition is the region under the restriction of \(f\) to the measurable set \(E \setminus E_{1}\), and
\[E_{1} \times [0,\infty) = \bigcup_{n=1}^{\infty} E_{1} \times [0,n]\]is a countable union of measurable sets.
Proposition (A function with finite integral is finite a.e.)
Let \(f : E \to [0,\infty]\) be measurable such that
\[\int_{E} f(x) \, dx < \infty.\]Then \(m(\{ x \in E : f(x) = \infty \}) = 0\), that is, \(f\) is finite almost everywhere.
Proof: Let \(E_{1} = \{ f = \infty \}\) and, for each \(n \in \mathbb{N}\), take
\[g(x) = n \mathbf{1}_{E_{1}}(x).\]Then \(g(x) \leq f(x)\) for \(x \in E\), since on \(E_{1}\) we have \(f = \infty \geq n\) and outside \(E_{1}\) we have \(g = 0 \leq f\). Hence, by the formula for simple functions and the monotonicity of the integral,
\[n\, m(E_{1}) = \int_{E} g(x)\, dx \leq \int_{E} f(x) \, dx < \infty\]for every \(n \in \mathbb{N}\), and therefore \(m(E_{1}) = 0\).
Decomposition of the domain and the monotone convergence theorem
Proposition (Additivity of the integral with respect to the domain)
Let \(f : E \to [0,\infty]\) be measurable. In the case where \(E = \bigcup_{i=1}^{\infty} E_{i}\) with the \(E_{i}\) measurable and \(E_{i} \cap E_{j} = \emptyset\) if \(i \neq j\), we have that
\[\int_{E} f(x) \, dx = \sum_{k=1}^{\infty} \int_{E_{k}} f(x) \, dx.\]This formula is also valid for finite unions.
Proof: Since the \(E_{i}\) are pairwise disjoint, we have that
\[R(f,E) = \bigcup_{k=1}^{\infty} R(f,E_{k}),\]and the \(R(f,E_{k})\) are pairwise disjoint, since their first coordinates live in disjoint sets. In this way, by countable additivity of the measure,
\[\int_{E} f(x) \, dx = m(R(f,E)) = m\left( \bigcup_{k=1}^{\infty} R(f,E_{k}) \right) = \sum_{k=1}^{\infty} m(R(f,E_{k})) = \sum_{k=1}^{\infty} \int_{E_{k}} f(x) \, dx.\]The formula for finite unions is obtained by taking \(E_{k} = \emptyset\) for the remaining indices, since \(R(f,\emptyset) = \emptyset\).
Theorem (Monotone convergence for non-negative functions)
Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of measurable functions such that
\[0 \leq f_{k}(x) \leq f_{k+1}(x)\]for every \(x \in E\). If
\[\lim_{k \to \infty} f_{k}(x) = f(x)\]almost everywhere on \(E\), then
\[\lim_{k \to \infty} \int_{E} f_{k}(x) \, dx = \int_{E} f(x) \, dx.\]Proof: Let \(Z \subseteq E\) be the null set where convergence fails, and \(\tilde{E} = E \setminus Z\). For any non-negative measurable function \(g\) we have \(R(g,Z) \subseteq Z \times [0,\infty) = \bigcup_{n} Z \times [0,n]\), which has measure zero; then \(\int_{Z} g \, dx = 0\) and, by additivity of the domain, the integrals over \(E\) and over \(\tilde{E}\) coincide both for \(f\) and for each \(f_{k}\). We may therefore suppose that \(f_{k}(x) \to f(x)\) for every \(x \in E\).
By the same approximation argument used for the region under the graph, we have that
\[R(f,E) = \Gamma(f,E) \cup \bigcup_{k=1}^{\infty} R(f_{k},E):\]if \(0 \leq y < f(x)\), since \(f_{k}(x) \uparrow f(x)\), there exists \(k\) with \(y \leq f_{k}(x)\), that is \((x,y) \in R(f_{k},E)\); if \(y = f(x) < \infty\), then \((x,y) \in \Gamma(f,E)\); and each \(R(f_{k},E) \subseteq R(f,E)\) because \(f_{k} \leq f\). Since
\[R(f_{k},E) \subseteq R(f_{k+1},E)\]by the monotonicity of the sequence, and \(m(\Gamma(f,E)) = 0\), continuity from below of the measure gives us
\[\int_{E} f(x) \, dx = m(R(f,E)) = m\left( \bigcup_{k=1}^{\infty} R(f_{k},E) \right) = \lim_{k \to \infty} m(R(f_{k},E)) = \lim_{k \to \infty} \int_{E} f_{k}(x) \, dx.\]A useful formula
Notation (Lower sum associated with a finite partition of the domain)
Given \(f : E \to [0,\infty]\) measurable and a finite partition \(E = \bigcup_{i=1}^{N} E_{i}\) into pairwise disjoint measurable sets, we define
\[\sum\big(f, \{E_{i}\}_{i=1}^{N}\big) = \sum_{i=1}^{N} \Big( \inf_{E_{i}} f \Big) \mathbf{1}_{E_{i}} \leq f.\]Theorem (The integral as a supremum of lower sums)
Let \(f : E \to [0,\infty]\) be measurable, non-negative. Then
\[\int_{E} f(x) \, dx = \sup \left\{ \int_{E} \sum\big(f, \{E_{i}\}_{i=1}^{N}\big) \, dx :\ E = \bigcup_{i=1}^{N} E_{i} \ \text{finite partition into disjoint measurable sets} \right\}.\]Proof: Given \(E_{1}, \dots, E_{N}\) pairwise disjoint measurable sets such that \(E = \bigcup_{i=1}^{N} E_{i}\), we have that
\[\sum\big(f, \{E_{i}\}_{i=1}^{N}\big) \leq f,\]then, by monotonicity and the formula for simple functions,
\[\sum_{i=1}^{N} \Big( \inf_{E_{i}} f \Big) m(E_{i}) = \int_{E} \sum\big(f, \{E_{i}\}_{i=1}^{N}\big) \, dx \leq \int_{E} f(x) \, dx.\]That is, the supremum in the statement is less than or equal to the integral.
For the reverse inequality, given \(k \geq 1\) and \(1 \leq i \leq k 2^{k}\), define
\[E_{k}^{0} = \{ f \geq k \}, \qquad E_{k}^{i} = \left\{ \frac{i-1}{2^{k}} \leq f(x) < \frac{i}{2^{k}} \right\}.\]Then
\[E = \bigcup_{i=0}^{k 2^{k}} E_{k}^{i}\]is a finite partition into pairwise disjoint measurable sets. Take
\[f_{k}(x) = k \mathbf{1}_{E_{k}^{0}} + \sum_{i=1}^{k 2^{k}} \left( \frac{i-1}{2^{k}} \right) \mathbf{1}_{E_{k}^{i}};\]this is the standard dyadic approximation, so we know that
\[f_{k} \leq f_{k+1} \qquad \text{and} \qquad \lim_{k \to \infty} f_{k} = f\]pointwise on \(E\). By the monotone convergence theorem,
\[\int_{E} f_{k} \, dx \to \int_{E} f \, dx.\]Since on each piece of the partition the value of \(f_{k}\) is less than or equal to the infimum of \(f\) on it, we have that
\[f_{k} \leq \sum\big(f, \{E_{k}^{i}\}_{i=0}^{k 2^{k}}\big) \leq f,\]and integrating,
\[\int_{E} f_{k} \, dx \leq \sum_{i=0}^{k 2^{k}} \Big( \inf_{E_{k}^{i}} f \Big) m(E_{k}^{i}) \leq \int_{E} f(x) \, dx.\]The middle term is one of the lower sums in the statement, and the left-hand side converges to \(\int_{E} f \, dx\); we then have that
\[\lim_{k \to \infty} \sum_{i=0}^{k 2^{k}} \Big( \inf_{E_{k}^{i}} f \Big) m(E_{k}^{i}) = \int_{E} f(x) \, dx,\]and the supremum in the statement reaches the integral.
Corollary (The integral as a supremum over simple functions)
Let \(f : E \to [0,\infty]\) be measurable. Then
\[\int_{E} f(x) \, dx = \sup \left\{ \int_{E} \phi(x) \, dx :\ \phi\ \text{simple and measurable},\ 0 \leq \phi \leq f\ \text{on } E \right\}.\]Proof: Exercise.
Null sets, Markov’s inequality and null functions
Lemma (The integral over a set of measure zero is zero)
Let \(f : E \to [0,\infty]\) be measurable. If \(m(E) = 0\), then
\[\int_{E} f(x) \, dx = 0.\]Proof: Let \(f_{k} = \sum_{i=1}^{m} a_{i} \mathbf{1}_{A_{i}}\), with \(A_{i} \cap A_{j} = \emptyset\) if \(i \neq j\), be non-negative simple functions increasing to \(f\), so that, by the monotone convergence theorem,
\[\lim_{k \to \infty} \int_{E} f_{k} \, dx = \int_{E} f(x) \, dx.\]Note that
\[\int_{E} f_{k}(x) \, dx = \sum_{i=1}^{m} a_{i}\, m(A_{i}).\]Since \(A_{i} \subseteq E\), we have \(m(A_{i}) = 0\) for \(1 \leq i \leq m\), and each \(\int_{E} f_{k} \, dx = 0\). We conclude that \(\int_{E} f \, dx = 0\).
Theorem (The integral respects a.e. inequalities)
Let \(f, g : E \to [0,\infty]\) be measurable such that \(g(x) \leq f(x)\) almost everywhere on \(E\). Then it holds that
\[\int_{E} g(x) \, dx \leq \int_{E} f(x) \, dx.\]In particular, if \(f = g\) a.e. on \(E\), then
\[\int_{E} g(x) \, dx = \int_{E} f(x) \, dx.\]Proof: Let
\[A = \{ x \in E :\ g(x) \leq f(x) \}.\]Then \(E = A \cup Z\) with \(Z = E \setminus A\) of measure zero and \(A \cap Z = \emptyset\). Thus, by additivity of the domain and the previous lemma, we have
\[\int_{E} f(x) \, dx = \int_{A} f(x) \, dx + \int_{Z} f(x) \, dx = \int_{A} f(x) \, dx,\] \[\int_{E} g(x) \, dx = \int_{A} g(x) \, dx + \int_{Z} g(x) \, dx = \int_{A} g(x) \, dx.\]Since \(g \leq f\) pointwise on \(A\), monotonicity of the integral gives \(\int_{A} g \, dx \leq \int_{A} f \, dx\), and the result follows. If moreover \(f = g\) a.e., both inequalities hold and equality is obtained.
Proposition (Markov’s inequality)
Let \(f : E \to [0,\infty]\) be measurable and \(\alpha > 0\). Then
\[m(\{ f \geq \alpha \}) \leq \frac{1}{\alpha} \int_{E} f(x) \, dx.\]Proof: Since \(f \geq 0\), we have the pointwise inequalities
\[\alpha \mathbf{1}_{\{f \geq \alpha\}} \leq f \mathbf{1}_{\{f \geq \alpha\}} \leq f.\]Integrating and using the formula for simple functions, the identity \(\int_{E} \mathbf{1}_{E_{1}} f \, dx = \int_{E_{1}} f \, dx\) and monotonicity,
\[\alpha\, m(\{ f \geq \alpha \}) \leq \int_{\{f \geq \alpha\}} f(x) \, dx \leq \int_{E} f(x) \, dx,\]and it suffices to divide by \(\alpha > 0\).
Corollary (A non-negative function with zero integral is zero a.e.)
Let \(f : E \to [0,\infty]\) be measurable such that \(\int_{E} f(x) \, dx = 0\). Then \(f = 0\) almost everywhere.
Proof: By Markov’s inequality, for every \(\alpha > 0\) we have that
\[m(\{ f \geq \alpha \}) \leq \frac{1}{\alpha} \int_{E} f(x) \, dx = 0.\]Hence
\[\{ f > 0 \} = \bigcup_{k=1}^{\infty} \left\{ f \geq \frac{1}{k} \right\}\]is a countable union of sets of measure zero, and therefore it is a set of measure zero. That is, \(f = 0\) a.e.
Linearity
Theorem (Linearity of the integral for non-negative functions)
Let \(f, g : E \to [0,\infty]\) be measurable and \(c \geq 0\). Then
\[\int_{E} \big( c f(x) + g(x) \big) \, dx = c \int_{E} f(x) \, dx + \int_{E} g(x) \, dx.\]Proof: Homogeneity. If \(c = 0\), with the convention \(0 \cdot \infty = 0\) we have \(cf \equiv 0\) and both sides of homogeneity are zero. So let \(c > 0\); we shall prove that
\[\int_{E} c f(x) \, dx = c \int_{E} f(x) \, dx.\]Let \(\{f_{k}\}\) be a sequence of non-negative simple functions such that
\[0 \leq f_{k} \leq f_{k+1} \qquad \text{and} \qquad \lim_{k \to \infty} f_{k} = f.\]If
\[f_{k} = \sum_{i=1}^{m_{k}} a_{i} \mathbf{1}_{A_{i}},\]with the \(A_{i}\) measurable and pairwise disjoint, we have that
\[\lim_{k \to \infty} c f_{k} = \lim_{k \to \infty} \sum_{i=1}^{m_{k}} c a_{i} \mathbf{1}_{A_{i}} = c f(x), \qquad c f_{k} \leq c f_{k+1}.\]Then, applying the monotone convergence theorem twice and the formula for simple functions,
\[\int_{E} c f(x) \, dx = \lim_{k \to \infty} \int_{E} c f_{k}(x) \, dx = \lim_{k \to \infty} \sum_{i=1}^{m_{k}} c a_{i}\, m(A_{i}) = c \lim_{k \to \infty} \sum_{i=1}^{m_{k}} a_{i}\, m(A_{i}) = c \int_{E} f(x) \, dx.\]Additivity. It remains to show additivity. To that end, let \(g_{k}\) be a sequence of non-negative simple functions such that
\[g_{k} \leq g_{k+1}, \qquad \lim_{k \to \infty} g_{k} = g.\]Then it follows that
\[g_{k} + f_{k} \leq g_{k+1} + f_{k+1}, \qquad \lim_{k \to \infty} (g_{k} + f_{k}) = g + f.\]We shall first show additivity for simple functions, that is, that
\[\int_{E} \big( g_{k}(x) + f_{k}(x) \big) \, dx = \int_{E} g_{k}(x) \, dx + \int_{E} f_{k}(x) \, dx.\]Let
\[g_{k} = \sum_{j=1}^{\ell_{k}} b_{j} \mathbf{1}_{B_{j}}, \quad \text{with } E = \bigcup_{j=1}^{\ell_{k}} B_{j} \text{ pairwise disjoint},\]and let us recall that
\[f_{k} = \sum_{i=1}^{m_{k}} a_{i} \mathbf{1}_{A_{i}}, \quad \text{with } E = \bigcup_{i=1}^{m_{k}} A_{i} \text{ pairwise disjoint}.\]Note that, since both families are partitions of \(E\),
\[\mathbf{1}_{A_{i}} = \sum_{j=1}^{\ell_{k}} \mathbf{1}_{B_{j} \cap A_{i}}, \qquad \mathbf{1}_{B_{j}} = \sum_{i=1}^{m_{k}} \mathbf{1}_{B_{j} \cap A_{i}},\]then
\[g_{k}(x) + f_{k}(x) = \sum_{j=1}^{\ell_{k}} \sum_{i=1}^{m_{k}} b_{j} \mathbf{1}_{A_{i} \cap B_{j}} + \sum_{i=1}^{m_{k}} \sum_{j=1}^{\ell_{k}} a_{i} \mathbf{1}_{A_{i} \cap B_{j}} = \sum_{j=1}^{\ell_{k}} \sum_{i=1}^{m_{k}} (b_{j} + a_{i}) \mathbf{1}_{A_{i} \cap B_{j}},\]which is a simple function on the common partition \(\{A_{i} \cap B_{j}\}\). Now, by finite additivity of the measure,
\[\sum_{i=1}^{m_{k}} m(A_{i} \cap B_{j}) = m\left( B_{j} \cap \bigcup_{i=1}^{m_{k}} A_{i} \right) = m(B_{j} \cap E) = m(B_{j}),\]and in the same way we have
\[\sum_{j=1}^{\ell_{k}} m(A_{i} \cap B_{j}) = m(A_{i}).\]Finally,
\[\begin{aligned} \int_{E} \big( f_{k}(x) + g_{k}(x) \big) \, dx &= \sum_{j=1}^{\ell_{k}} \sum_{i=1}^{m_{k}} (a_{i} + b_{j})\, m(A_{i} \cap B_{j}) \\ &= \sum_{j=1}^{\ell_{k}} b_{j} \sum_{i=1}^{m_{k}} m(A_{i} \cap B_{j}) + \sum_{i=1}^{m_{k}} a_{i} \sum_{j=1}^{\ell_{k}} m(A_{i} \cap B_{j}) \\ &= \sum_{j=1}^{\ell_{k}} b_{j}\, m(B_{j}) + \sum_{i=1}^{m_{k}} a_{i}\, m(A_{i}) \\ &= \int_{E} g_{k}(x) \, dx + \int_{E} f_{k}(x) \, dx. \end{aligned}\]Since \(\{f_{k} + g_{k}\}\) is an increasing sequence of simple functions converging to \(f + g\), the monotone convergence theorem applied three times gives us
\[\int_{E} (f + g) \, dx = \lim_{k \to \infty} \int_{E} (f_{k} + g_{k}) \, dx = \lim_{k \to \infty} \left( \int_{E} f_{k} \, dx + \int_{E} g_{k} \, dx \right) = \int_{E} f \, dx + \int_{E} g \, dx.\]Combining homogeneity with additivity, applied to \(cf\) and \(g\), we obtain the statement.