9. Measurable functions PDF TEX
Measurable functions
Induced mapping
Any mapping \(f:X \to Y\) between two sets \(X\) and \(Y\) induces a mapping \(f^{-1}:\mathcal{P}(Y) \to \mathcal{P}(X)\) defined by \[f^{-1}[E]=\{x \in X: f(x) \in E\}\subseteq X \quad \text{ for every } \quad E \subseteq Y.\] This mapping preserves unions, intersections, and complements, i.e. \[\begin{aligned} f\Big[\bigcup_{\alpha\in A}E_\alpha\Big]&=\bigcup_{\alpha\in A}f[E_{\alpha}],\\ f\Big[\bigcap_{\alpha\in A}E_\alpha\Big]&=\bigcap_{\alpha\in A}f[E_{\alpha}],\\ f[E^c]&=f[E]^c. \end{aligned}\]If \(\mathcal{N}\) is a \(\sigma\)-algebra on \(Y\), then \(\{f^{-1}[E]: E \in \mathcal{N}\}\) is a \(\sigma\)-algebra on \(X\).
Measurable mapping
If \((X,\mathcal M)\) and \((Y,\mathcal N)\) are measurable spaces, a mapping \(f:X \to Y\) is called \((\mathcal M,\mathcal N)\)-measurable or just measurable when \(\mathcal M\) and \(\mathcal N\) are understood, if \[f^{-1}[E] \in \mathcal M \quad \text{ for all } \quad E \in \mathcal N.\]
Remark. The composition of measurable mappings is measurable. More precisely, if \(f:X \to Y\) is \((\mathcal M,\mathcal N)\)-measurable and \(g:Y \to Z\) is \((\mathcal N,\mathcal O)\)-measurable then \[g \circ f:X \to Z\] is \((\mathcal M,\mathcal O)\)-measurable.
Useful facts
Proposition. If \(\mathcal N\) is generated by \(\mathcal{E}\), then \(f:X \to Y\) is \((\mathcal M,\mathcal N)\)-measurable iff \[f^{-1}[E] \in \mathcal M\quad \text{ for all } \quad E \in \mathcal{E}.\]
Proof (\(\Longrightarrow\)). If \(f:X \to Y\) is measurable then obviously \(f^{-1}[E] \in \mathcal M\) for all \(E \in \mathcal{E}\) since \(\mathcal N=\sigma(\mathcal{E}) \supseteq \mathcal{E}\).
Proof (\(\Longleftarrow\)). Observe that \(\mathcal O=\{E \subseteq Y: f^{-1}[E] \in \mathcal M\}\) is a \(\sigma\)-algebra that contains \(\mathcal{E}\) thus \(\mathcal N =\sigma(\mathcal{E}) \subseteq \mathcal O\).$$\tag*{$\blacksquare$}$$
If \(X\) and \(Y\) are metric or topological spaces, every continuous function \(f:X \to Y\) is \((\mathcal{B}_X,\mathcal{B}_Y)\)-measurable. Here we abbreviate \({\rm Bor}(Z)\) to \(\mathcal{B}_Z\).
Proof. \(f\) is continuous iff \(f^{-1}[U]\) is open in \(X\) for every open \(U\subseteq Y\).$$\tag*{$\blacksquare$}$$
Measurable functions
If \((X, \mathcal M)\) is measurable space, a real-valued or complex-valued function \(f\) on \(X\) will be called \(\mathcal M\)-measurable or just measurable if it is \((\mathcal M,\mathcal{B}_\mathbb{R})\) or \((\mathcal M,\mathcal{B}_\mathbb{C})\) measurable.
In particular, \(f:\mathbb{R} \to \mathbb{C}\) is Lebesgue (resp. Borel) measurable if it is \((\mathcal{L}(\mathbb R),\mathcal{M}_{\mathbb{C}})\) (resp. \((\mathcal{B}_{\mathbb{R}},\mathcal{B}_{\mathbb{C}})\)) measurable, likewise for \(f:\mathbb{R} \to \mathbb{R}\).
Remark. (Warning!) If \(f,g:\mathbb{R} \to \mathbb{R}\) are Lebesgue measurable it does not follow that \(f \circ g\) is Lebesgue measurable, even if \(g\) is assumed continuous.
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If \(E \in \mathcal{B}_{\mathbb{R}}\) we have \(f^{-1}[E] \in \mathcal{L}(\mathbb R)\), but unless \(f^{-1}[E] \in \mathcal{B}_{\mathbb{R}}\) there is no guarantee that \(g^{-1}\big[f^{-1}[E]\big]\) will be in \(\mathcal{L}(\mathbb R)\).
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However, is \(f\) is Borel measurable then \(f \circ g\) is Lebesgue or Borel whenever \(g\) is.
Proposition
Proposition. If \((X,\mathcal M)\) is a measurable space and \(f:X \to \mathbb{R}\) the following are equivalent:
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\(f\) is \(\mathcal M\)-measurable,
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\(f^{-1}[(a,\infty)] \in \mathcal M\) for all \(a \in \mathbb{R}\),
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\(f^{-1}[[a,\infty)] \in \mathcal M\) for all \(a \in \mathbb{R}\),
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\(f^{-1}[(-\infty,a)] \in \mathcal M\) for all \(a \in \mathbb{R}\),
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\(f^{-1}[(-\infty,a]] \in \mathcal M\) for all \(a \in \mathbb{R}\).
Proof. It follows from the previous proposition and the fact that \[\begin{aligned} \mathcal{B}_{\mathbb{R}}&=\sigma(\{(a,\infty): a \in \mathbb{R}\})=\sigma(\{[a,\infty): a \in \mathbb{R}\})\\ &=\sigma(\{(-\infty, a): a \in \mathbb{R}\}) =\sigma(\{(-\infty, a]: a \in \mathbb{R}\}). \qquad {\blacksquare} \end{aligned}\]
Measurable functions in products
Given a set \(X\), if \(\{(Y_{\alpha},\mathcal{N}_{\alpha}): \alpha \in A\}\) is a family of measurable spaces and \(f_{\alpha}:X \to Y_{\alpha}\) is a map for each \(\alpha \in A\), there is a unique smallest \(\sigma\)-algebra on \(X\) with respect to which the \(f_{\alpha}\)’s are all measurable, namely the \(\sigma\)-algebra generated by the family \[\{f^{-1}_{\alpha}[E_{\alpha}]: E_{\alpha} \in \mathcal{N}_{\alpha} \text{ and }\alpha \in A\}.\] It is called the \(\sigma\)-algebra generated by the family \(\{f_{\alpha}:\alpha \in A\}\).
Example. If \(X=\prod_{\alpha \in A}Y_{\alpha}\), then \[\bigotimes_{\alpha \in A}\mathcal{N}_{\alpha}=\sigma\big(\{\pi_{\alpha}^{-1}[E_{\alpha}]: E_{\alpha} \in \mathcal{N}_{\alpha} \text{ and } \alpha \in A\}\big)\] is the \(\sigma\)-algebra generated by the coordinate maps \(\pi_{\alpha}:X \to Y_{\alpha}\).
Useful proposition
Proposition. Let \((X,\mathcal M)\) and \(\{(Y_{\alpha},\mathcal{N}_{\alpha}): \alpha \in A\}\) be measurable spaces, \(Y=\prod_{\alpha \in A}Y_{\alpha}\), \[\mathcal N=\bigotimes_{\alpha \in A}\mathcal N_{\alpha}= \sigma\big(\{\pi_{\alpha}^{-1}[E_{\alpha}]: E_{\alpha}\in \mathcal N_{\alpha} \text{ and } \alpha \in A\}\big),\] where \(\pi_{\alpha}:Y \to Y_{\alpha}\) is the coordinate map for \(\alpha\in A\). Then \(f:X \to Y\) is \((\mathcal M,\mathcal N)\)-measurable iff \(f_{\alpha}=\pi_{\alpha} \circ f\) is \((\mathcal M,\mathcal N_{\alpha})\)-measurable for all \(\alpha \in A\).
In particular, \(f:X \to \prod_{j=1}^{\infty}X_j\) given by \(f(x)=(f_1(x),f_2(x),f_3(x),\ldots)\) is measurable iff all \(f_j\)’s are measurable.
Proof (\(\Longrightarrow\)). If \(f:X \to Y\) is \((\mathcal M,\mathcal N)\)-measurable and \(\pi_{\alpha}: Y\to Y_{\alpha}\) is \((\mathcal N,\mathcal N_{\alpha})\)-measurable then \(f_{\alpha}=\pi_{\alpha} \circ f\) is \((\mathcal M,\mathcal N_{\alpha})\)-measurable.
Proof (\(\Longleftarrow\)). If each \(f_{\alpha}\) is \((\mathcal M,\mathcal N_{\alpha})\)-measurable then for all \(E_{\alpha} \in \mathcal N_{\alpha}\) we have \(f^{-1}[\pi_{\alpha}^{-1}[E_{\alpha}]]=f^{-1}_{\alpha}[E_{\alpha}] \in \mathcal M\), so \(f\) is measurable by the previous proposition.$$\tag*{$\blacksquare$}$$
Corollary
Let \((X,\mathcal M)\) be a measurable space. A function \(f:X \to \mathbb{C}\) is \(\mathcal M\)-measurable iff \({\rm Re}(f)\) and \({\rm Im}(f)\) are \(\mathcal M\)-measurable.
Proof. This follows from the fact that \[\mathcal{B}_{\mathbb{C}}=\mathcal{B}_{\mathbb{R}^2}=\mathcal{B}_{\mathbb{R}}\otimes \mathcal{B}_{\mathbb{R}}.\] Moreover, we have \[f(x)={\rm Re}(f)(x)+i\:{\rm Im}(f)(x) \sim ({\rm Re}(f)(x),{\rm Im}(f)(x)).\] This completes the proof. $$\tag*{$\blacksquare$}$$
Extended real numbers system
If \((X,\mathcal M)\) is a measurable space, \(f\) is a function on \(X\) and \(E \in \mathcal{M}\) we say \(f\) is measurable on \(E\) if \(f^{-1}[B] \cap E \in \mathcal M\) for all Borel sets \(B\in \mathcal B_X\).
Remark. Let \(\overline{\mathbb{R}}=[-\infty,\infty]\) be the extended real number system. We define Borel sets in \(\overline{\mathbb{R}}\) by \[\mathcal{B}_{\overline{\mathbb{R}}}=\{E \subseteq \overline{\mathbb{R}}: E \cap \mathbb{R} \in \mathcal{B}_{\mathbb{R}}\}.\] It can be easily verified that \[\mathcal{B}_{\overline{\mathbb{R}}}=\sigma\left(\{(a,\infty]:a \in \mathbb{R}\}\right)=\sigma\left(\{[-\infty,a): a \in \mathbb{R}\}\right),\] and we define \(f:X \to \overline{\mathbb{R}}\) to be \(\mathcal M\)-measurable if it is \((\mathcal M,\mathcal{B}_{\overline{\mathbb{R}}})\)-measurable.
Mesurability is preserved under algebraic operations
Proposition. Let \((X,\mathcal M)\) be a measurable space. If \(f,g:X \to \mathbb{C}\) are \(\mathcal M\)-measurable, then so are \(f+g\) and \(f \cdot g\).
Proof. Define \(F:X \to \mathbb{C} \times \mathbb{C}\), \(\phi:\mathbb{C} \times \mathbb{C} \to \mathbb{C}\) and \(\psi:\mathbb{C} \times \mathbb{C} \to \mathbb{C}\) by \[F(x)=(f(x),g(x)), \qquad \phi(z,w)=z+w, \qquad \psi(z,w)=z\cdot w.\] Since \[\mathcal{B}_{\mathbb{C} \times \mathbb{C}}=\mathcal{B}_{\mathbb{C}} \otimes \mathcal{B}_{\mathbb{C}},\] \(F\) is \((\mathcal M,\mathcal{B}_{\mathbb{C} \times \mathbb{C}})\)-measurable by the previous proposition. The functions \(\phi\) and \(\psi\) are \((\mathcal{B}_{\mathbb{C} \times \mathbb{C}}, \mathcal{B}_{\mathbb{C}})\)-measurable, since \(\phi\), \(\psi\) are continuous. Thus \[{\color{red}f+g=\phi \circ F,} \quad \text{ and } \quad {\color{blue}f \cdot g=\psi \circ F}\] are \(\mathcal M\)-measurable.$$\tag*{$\blacksquare$}$$
Mesurability is preserved under limiting operations
Proposition. Let \((X,\mathcal M)\) be a measurable space. If \((f_n)_{n \in \mathbb{N}}\) is a sequence of \(\overline{\mathbb{R}}\)-valued measurable functions, then the functions \[\begin{aligned} g_1(x)&=\sup_{n\in\mathbb N}f_n(x), \quad g_3(x)=\limsup_{n \to \infty}f_n(x),\\ g_2(x)&=\inf_{n\in\mathbb N}f_n(x), \quad g_4(x)=\liminf_{n \to \infty}f_n(x), \end{aligned}\] are all measurable. If \[f(x)=\lim_{n \to \infty}f_n(x)\] exists for every \(x \in X\), then \(f\) is measurable.
If \(f,g:X \to \overline{\mathbb{R}}\) are measurable, then so are \(\max(f,g)\) and \(\min(f,g)\).
If \((f_n)_{n \in \mathbb{N}}\) is a sequence of complex-valued measurable functions and \[f(x)=\lim_{n \to \infty}f_n(x)\] exists for all \(x \in X\), then \(f\) is measurable.
Positive \(f^{+}\) and negative \(f^{-}\) parts of \(f\)
If \(f:X \to \overline{\mathbb{R}}\) we define the positive \(f^{+}\) and negative \(f^{-}\) parts of \(f\) to be \[{\color{blue}f^{+}(x)=\max(f(x),0)}, \quad \text{ and } \quad {\color{red}f^{-}(x)=\max(-f(x),0)}.\] Then \(f=f^{+}-f^{-}\). If \(f\) is measurable, so are \(f^{+}\) and \(f^{-}\).
Polar decomposition
If \(f:X \to \mathbb{C}\), we have its polar decomposition \[f={\rm sgn}(f)|f|,\] where \[{\rm sgn}(z)=\begin{cases} \frac{z}{|z|} &\text{ if }z \neq 0,\\\ 0 &\text{ otherwise.} \end{cases}\]
Remark. If \(f\) is measurable, so are \(|f|\) and \({\rm sgn}(f)\).
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If \(f\) is measurable, so are \(|f|\) and \({\rm sgn}(f)\). Indeed, \(z \mapsto |z|\) is continuous on \(\mathbb{C}\), and \(z \mapsto {\rm sgn}(x)\) is continuous except at the origin.
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If \(U \subseteq \mathbb{C}\) is open \({\rm sgn}^{-1}[U]\) is either open or of the form \(V \cup \{0\}\), where \(V\) is open, so \({\rm sgn}\) is Borel measurable.
Characteristic functions and simple functions
Suppose that \((X,\mathcal M)\) is a measurable space. If \(E \subseteq X\), the characteristic function of \(E\) (sometimes called the indicator function of \(E\)) is defined by \[\mathbf{1}_{{E}}(x)=\begin{cases} 1 &\text{ if }x \in E,\\ 0 &\text{ if }x \not\in E. \end{cases}\] \(\mathbf{1}_{{E}}\) is measurable iff \(E \in \mathcal M\).
A simple function on \(X\) is a finite linear combination, with complex coefficients, of characteristic functions of sets in \(\mathcal M\). We do not allow simple functions to assume the values \(\pm\infty\). Equivalently, \(f:X \to \mathbb{C}\) is simple iff \(f\) is measurable and the range of \(f\) is a finite subset of \(\mathbb{C}\).
Standard representation of simple functions
Definition. Assume that the range of \(f\) is \(\{z_1,\ldots, z_n\}\subset \mathbb C\). Then we can write \[f=\sum_{j=1}^n z_j\mathbf{1}_{{E_j}},\quad \text{ where } \quad E_j=f^{-1}[\{z_j\}].\] This is called the standard representation of \(f\). It exhibits \(f\) a linear combination, with distinct coefficients, of characteristic functions of disjoint sets whose union is \(X\), i.e. \[E_j \cap E_i=\varnothing \quad \text{ if } \quad i \neq j, \quad \bigcup_{j=1}^{n}E_j=X.\]
Remark. If \(f\) and \(g\) are simple functions, then so are \(f+g\) and \(f \cdot g\).
Measurable functions are limits of simple functions
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If \(f:X \to [0,\infty]\) is measurable then there is a sequence \((\phi_n)_{n \in \mathbb{N}}\) of simple functions such that \(0 \leq f_1 \leq f_2\le \ldots \leq f\) and \(f_n \ _{\overrightarrow{n\to\infty}}\ f\) pointwise, and \(f_n \ _{\overrightarrow{n\to\infty}}\ f\) uniformly on any set on which \(f\) is bounded.
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If \(f:X \to \mathbb{C}\) is measurable then there is a sequence \((f_n)_{n \in \mathbb{N}}\) of simple functions such that \(0 \leq |f_1| \leq |f_2| \leq \ldots \leq |f|\) and \(f_n \ _{\overrightarrow{n\to\infty}}\ f\) pointwise, and \(f_n \ _{\overrightarrow{n\to\infty}}\ f\) uniformly on any set on which \(f\) is bounded.
Proof of (a). For \(n\in \{0,1,2,\ldots\}\) and \(0 \leq k \leq 2^{2n}-1\), let \[\begin{gathered} E_n^k=f^{-1}\big[(k2^{-n},(k+1)2^{-n}]\big]\quad \text{ and } \quad F_n=f^{-1}\big[(2^n,\infty]\big],\\ f_n(x)=\sum_{k=0}^{2^{2n}-1}k2^{-n}\mathbf{1}_{{E_n^k}}(x)+2^n\mathbf{1}_{{F_n}}(x). \end{gathered}\] For \(n\in \mathbb N\), one can easily see \[f_n(x) \leq f_{n+1}(x) \leq f(x).\]
Moreover, on the set \(F_n^c\) (since \(f(x) \leq 2^n\)), we have \[\begin{aligned} 0 &\leq f(x)-f_n(x)= \sum_{k=0}^{2^{2n}-1}(f(x)-k2^{-n})\mathbf{1}_{{E_n^k}}(x)\le \sum_{k=0}^{2^{2n}-1}{2^{-n}}\mathbf{1}_{{E_n^k}}(x) \leq {2^{-n}}. \end{aligned}\]
Proof of (b). If \(f=g+ih\) we apply part (a) to the positive and negative parts of \(g\) and \(h\) obtaining sequences \[g_n^{+} \ _{\overrightarrow{n\to\infty}}\ g^{+}, \qquad g_n^{-} \ _{\overrightarrow{n\to\infty}}\ g^{-},\] \[h_n^{+} \ _{\overrightarrow{n\to\infty}}\ h^{+}, \qquad h_n^{-} \ _{\overrightarrow{n\to\infty}}\ h^{-},\] of nonnegative increasing simple functions. Taking \[f_n=g_n^+-g_n^{-}+i(h_n^{+}-h_n^{-}),\] we can easily verify that \((f_n)_{n\in\mathbb N}\) has the desired properties. $$\tag*{$\blacksquare$}$$
Proposition
Proposition. Let \((X, \mathcal M, \mu)\) be a complete measure space. Then the following are true.
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If \(f\) is measurable and \(f=g\) \(\mu\)-a.e., then \(g\) is measurable.
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If \((f_n)_{n\in\mathbb N}\) is a sequence of measurable functions and \(f_n \ _{\overrightarrow{n\to\infty}}\ f\) \(\mu\)-a.e., then \(f\) is measurable.
Remark.
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If \((f_n)_{n\in\mathbb N}\) is a sequence of measurable functions, then the set \(D_e=\{x \in X: \lim_{n \to \infty}f_n(x) \text{ exists}\}\) is measurable, and \(\lim_{n \to \infty}f_n(x)\) is measurable on \(D_e\).
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If \(\mu\) is not complete and \(f_n \ _{\overrightarrow{n\to\infty}}\ f\) \(\mu\)-a.e., then one can perturb \(f\) on the set \[D_{ne}=\{x \in X: \lim_{n \to \infty}f_n(x) \text{ does not exist}\},\] in a way that \(f\) is not measurable on \(X\).
Proposition
Proposition. Let \((X,\mathcal M,\mu)\) be a measurable space and let \((X,\overline{\mathcal M},\overline{\mu})\) be its completion. If \(f\) is \(\overline{\mathcal M}\)-measurable function on \(X\), there is \(\mathcal M\)-measurable function \(g\) such that \(f=g\) \(\overline{\mu}\)-a.e.
Proof. From the definition of \(\overline{\mu}\) it is obvious if \(f=\mathbf{1}_{{E}}\), where \(E\in\mathcal M\), and hence if \(f\) is an \(\overline{\mathcal M}\)-measurable simple function.
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For the general case, choose a sequence \((f_n)_{n\in\mathbb N}\) of \(\overline{\mathcal M}\)-measurable simple functions that converge pointwise to \(f\) by the previous theorem.
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For each \(n\in \mathbb N\) let \(g_n\) be an \(\mathcal M\)-measurable simple function with \(f_n=g_n\) except on a set \(E_n\in \overline{\mathcal M}\) with \(\overline{\mu}(E_n)=0\).
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Choose \(N\in \mathcal M\) such that \(\mu(N)=0\) and \(N\supseteq \bigcup_{n\in\mathbb N}E_n\) and set \(g=\lim_{n\to \infty}\mathbf{1}_{{N^c}}g_n\). Then \(g\) is \(\mathcal M\)-measurable and \(g=f\) on \(N^c\). $$\tag*{$\blacksquare$}$$