7. 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. \[\begin{align*} f\Big[\bigcup_{\alpha\in A}E_\alpha\Big]&=\bigcup_{\alpha\in A}f[E_{\alpha}],\\ f\Big[\bigcap_{\alpha\in A}E_\alpha\Big]&=\bigcap_{\alpha\in A}f[E_{\alpha}],\\ f[E^c]&=f[E]^c. \end{align*}\]
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 1. 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 \(\mathcal N =\sigma(\mathcal{E}) \subseteq \mathcal O\).
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 \(U\subseteq Y\).
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.
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 \(\mathcal{L}(\mathbb R)\).
However, is \(f\) is Borel measurable then \(f \circ g\) is Lebesgue or Borel whenever \(g\) is.
Proposition
Proposition 1. If \((X,\mathcal M)\) is a measurable space and \(f:X \to \mathbb{R}\) the following are equivalent:
\(f\) is \(\mathcal M\)-measurable,
\(f^{-1}[(a,\infty)] \in \mathcal M\) for all \(a \in \mathbb{R}\),
\(f^{-1}[[a,\infty)] \in \mathcal M\) for all \(a \in \mathbb{R}\),
\(f^{-1}[(-\infty,a)] \in \mathcal M\) for all \(a \in \mathbb{R}\),
\(f^{-1}[(-\infty,a]] \in \mathcal M\) for all \(a \in \mathbb{R}\).
Proof. It follows from the previous proposition and the fact that \[\begin{align*} \mathcal{B}_{\mathbb{R}}&=\sigma(\{(a,\infty): a \in \mathbb{R}\})=\sigma(\{[a,\infty): a \in \mathbb{R}\})\\ &=\sigma(\{(-\infty, a): a \in \mathbb{R}\}) =\sigma(\{(-\infty, a]: a \in \mathbb{R}\}). \qquad \end{align*}\]
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 1. 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.
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.
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 1. 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.
Mesurability is preserved under limiting operations
Proposition 1. 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 \[\begin{align*} g_1(x)&=\sup_{n\in\mathbb N}f_n(x), \quad g_3(x)=\limsup_{n \to \infty}f_n(x),\\ g_2(x)&=\inf_{n\in\mathbb N}f_n(x), \quad g_4(x)=\liminf_{n \to \infty}f_n(x), \end{align*}\] are all measurable. If \[f(x)=\lim_{n \to \infty}f_n(x)\] exists for every \(x \in X\), then \(f\) is measurable.
We have to show \(g_1^{-1}((a,\infty]) \in \mathcal M\) for all \(a \in \mathbb{R}\). \[\begin{align*} g_1^{-1}\big[(a,\infty]\big]&=\{x \in X : g_1(x) \in (a,\infty]\}\\ &=\{x \in X : \sup_{n\in\mathbb N}f_n(x) >a\}\\ &=\bigcup_{n\in\mathbb N}\{x \in X: f_n(x) >a\}\\ &=\bigcup_{n\in\mathbb N}f_n^{-1}\big[(a,\infty]\big] \in \mathcal M. \end{align*}\] Similarly \[g_2^{-1}\big[[-\infty,a)\big]=\bigcup_{n\in\mathbb N}f_n^{-1}\big[[-\infty,a)\big] \in \mathcal M.\] Thus \(g_1\), \(g_2\) are measurable.
Observe that \[\begin{align*} g_3(x)=\limsup_{n \to \infty}f_n(x)=\lim_{k \to \infty}\sup_{n>k}f_n(x)=\lim_{k \to \infty}h_k(x), \end{align*}\] where \(h_k(x)=\sup_{n>k}f_n(x)\), and we have \[h_{k+1}(x) \leq h_k(x), \quad \text{ thus } \quad \lim_{k \to \infty}h_k(x)=\inf_{k\in\mathbb N}h_k(x).\] From the previous slide \(h_k(x)=\sup_{n>k}f_n(x)\) is \(\mathcal M\)-measurable and \(\inf_{k\in \mathbb N}h_k(x)\) is also \(\mathcal M\)-measurable. Thus \[g_3(x)=\limsup_{n \to \infty}f_n(x)\] is \(\mathcal M\)-measurable, and likewise for \(g_4\). Finally, if \(f(x)=\lim_{n \to \infty}f_n(x)\) exists then \(f(x)=g_3(x)=g_4(x)\), so \(f\) is \(\mathcal M\)-measurable.
Two corollaries
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 \({\rm sgn}(f)\).
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.
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
Standard representation of simple functions. 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
Theorem. Let \((X,\mathcal M)\) be a measurable space.
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.
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 \[\begin{gather*} E_n^k=f^{-1}\big[(k2^{-n},(k+1)2^{-n}]\big]\quad \text{ and } \quad F_n=f^{-1}\big[(2^n,\infty]\big],\\ f_n(x)=\sum_{k=0}^{2^{2n}-1}k2^{-n}\mathbf{1}_{{E_n^k}}(x)+2^n\mathbf{1}_{{F_n}}(x). \end{gather*}\] 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 \[\begin{align*} 0 &\leq f(x)-f_n(x)= \sum_{k=0}^{2^{2n}-1}(f(x)-k2^{-n})\mathbf{1}_{{E_n^k}}(x)\le \sum_{k=0}^{2^{2n}-1}{2^{-n}}\mathbf{1}_{{E_n^k}}(x) \leq {2^{-n}}. \end{align*}\]
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.
Proposition
Proposition 1. Let \((X, \mathcal M, \mu)\) be a complete measure space. Then the following are true.
If \(f\) is measurable and \(f=g\) \(\mu\)-a.e., then \(g\) is measurable.
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.
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 \(D_e\).
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 1. 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.
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.
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 \(\overline{\mu}(E_n)=0\).
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 \(N^c\).
Problems
Completeness and measurability
Problem 1. Let \((X, \mathcal M, \mu)\) be a measure space. Show that each of the following implications is valid if and only if the measure \(\mu\) is complete:
if \(f\) is measurable and \(f=g\) \(\mu\)-a.e., then \(g\) is measurable;
if \(f_n\) is measurable for every \(n \in \mathbb N\) and \(f_n \ _{\overrightarrow{n \to \infty}}\ f\) \(\mu\)-a.e., then \(f\) is measurable.
One implication is the proposition proved above. For the converse take a set \(N \in \mathcal M\) with \(\mu(N)=0\) and \(A \subseteq N\), and consider the functions \(f=0\) and \(g=\mathbf{1}_{{A}}\).
Continuous images of null sets
Problem 2. Show that there exist a strictly increasing continuous function \(f\) on \([0,1]\) and a compact set \(K \subseteq [0,1]\) such that \[\lambda_1(K)=0 \quad \text{ and } \quad \lambda_1(f[K])=1.\]
Hint: let \(c\) be the Cantor–Lebesgue function from Lecture 6, let \(K\) be the Cantor set and consider \(f(x)=x+c(x)\).
In particular, continuous functions need not map null sets to null sets, and therefore they need not map Lebesgue measurable sets to Lebesgue measurable sets.
Continuous images and preimages of measurable sets
Problem 3. Show that there exist a set \(E \in \mathcal L(\mathbb R)\) and a continuous function \(f\) on an interval containing \(E\) such that \[f[E] \not\in \mathcal L(\mathbb R).\] Hint: with \(f\) and \(K\) as in Problem 2, the set \(f[K]\) has positive measure, so it contains a nonmeasurable set \(A\); consider \(E=f^{-1}[A] \cap K\).
Problem 4. Show that there exists a real-valued \(\mathcal L(\mathbb R)\)-measurable function \(g\) for which \[g^{-1}[E] \not\in \mathcal L(\mathbb R) \quad \text{ for some } \quad E \in \mathcal L(\mathbb R).\] Hint: take \(g=f^{-1}\), where \(f\) is the function from Problem 2, and \(E\) from Problem 3.
Lebesgue measurable sets which are not Borel
Problem 5. Show that there exists a set \(E \in \mathcal L(\mathbb R)\) such that \[E \not\in {\rm Bor}(\mathbb R).\] Hint: the set \(E\) from Problem 3 works. If \(E\) were a Borel set, then \(f[E]\) would be a Borel set as well, since \(f\) is a homeomorphism of \([0,1]\) onto \(f[[0,1]]\).
Compare with Lecture 6, where the same conclusion was obtained by a cardinality argument. The present proof is constructive and, in addition, exhibits a null set which is not Borel.