Closure properties
Let \((X,\mathcal M)\) be a measurable space and \(f, g, f_1, f_2,\ldots\) measurable functions \(X\to\RR\). Then the following are measurable:
\(\varphi\circ f\) for continuous \(\varphi\colon\RR\to\RR\);
\(\lim_n f_n\) where it exists, and \(\liminf_n f_n\), \(\limsup_n f_n\), \(\sup_n f_n\), \(\inf_n f_n\);
\(f+g\), \(fg\), and \(f/g\) where \(g\) does not vanish;
a function \(h\) such that \(h|_{E_k}\) is measurable for each set \(E_k\) in a countable measurable cover \(\theset{E_k}\) of \(X\).
Continuous functions and monotone functions \(\RR\to\RR\) are Borel measurable, and a function equal almost everywhere to a Lebesgue measurable function is Lebesgue measurable.
Order of composition
If \(\varphi\) is continuous and \(f\) is Lebesgue measurable, then \(\varphi\circ f\) is Lebesgue measurable, since \((\varphi\circ f)^{-1}(U) = f^{-1}(\varphi^{-1}(U))\) and \(\varphi^{-1}(U)\) is open. A composition \(f\circ\varphi\) with \(f\) Lebesgue measurable and \(\varphi\) continuous need not be Lebesgue measurable, because \(\varphi^{-1}\) of a Lebesgue measurable set need not be Lebesgue measurable; the homeomorphism \(x\mapsto x+c(x)\), for \(c\) the Cantor function, maps the Cantor set onto a set of measure \(1\) and gives an example.
A composition of Lebesgue measurable functions need not be measurable (Tag FF-UW3C7)
There exist a Lebesgue measurable function \(h\colon\RR\to\RR\) and a continuous function \(\phi\colon[0,2]\to[0,1]\) such that \(h\circ\phi\) is not Lebesgue measurable.
Let \(C\subseteq[0,1]\) be the Cantor set and \(f\colon[0,1]\to[0,1]\) the Cantor–Lebesgue function, which is continuous, nondecreasing, and constant on each connected component of \([0,1]\setminus C\). Put \(g(x)\coloneqq f(x) + x\). Then \(g\colon[0,1]\to[0,2]\) is continuous and strictly increasing with \(g(0) = 0\) and \(g(1) = 2\), so it is a homeomorphism, and \(\phi\coloneqq \inverseof{g}\colon[0,2]\to[0,1]\) is continuous.
The set \([0,1]\setminus C\) is a disjoint union of open intervals of total length \(1\), and \(g\) maps each of them onto an interval of the same length because \(f\) is constant on it. Hence \(m(g([0,1]\setminus C)) = 1\) and \(m(g(C)) = 2 - 1 = 1 > 0\). Every subset of \(\RR\) of positive Lebesgue measure contains a subset that is not Lebesgue measurable, so there is a non-measurable \(A\subseteq g(C)\).
The set \(B\coloneqq \inverseof{g}(A)\) is contained in \(C\), which has measure \(0\), so \(B\) is Lebesgue measurable and \(h\coloneqq\chi_B\) is a Lebesgue measurable function. For \(y\in[0,2]\), \((h\circ\phi)(y) = 1\) if and only if \(\inverseof{g}(y)\in \inverseof{g}(A)\), that is, \(y\in A\). Hence \(\inverseof{(h\circ\phi)}(\theset{1}) = A\) is not Lebesgue measurable, so \(h\circ\phi\) is not Lebesgue measurable.
A composition of Lebesgue measurable functions need not be measurable (Tag FE-YOIJM)
There exist a Lebesgue measurable function \(h\colon\RR\to\RR\) and a continuous function \(\phi\colon[0,2]\to[0,1]\) such that \(h\circ\phi\) is not Lebesgue measurable.
Let \(C\subseteq[0,1]\) be the Cantor set and \(f\colon[0,1]\to[0,1]\) the Cantor–Lebesgue function, which is continuous, nondecreasing, and constant on each connected component of \([0,1]\setminus C\). Put \(g(x)\coloneqq f(x) + x\).
The map \(g\colon[0,1]\to[0,2]\) is continuous and strictly increasing with \(g(0) = 0\) and \(g(1) = 2\), so it is a homeomorphism, and \(\phi\coloneqq \inverseof{g}\colon[0,2]\to[0,1]\) is continuous.
The set \([0,1]\setminus C\) is a disjoint union of open intervals of total length \(1\). On each such interval \(f\) is constant, so \(g\) maps it onto an interval of the same length. Hence \(m(g([0,1]\setminus C)) = 1\) and \(m(g(C)) = m([0,2]) - 1 = 1 > 0\).
Every subset of \(\RR\) of positive Lebesgue measure contains a subset that is not Lebesgue measurable, so there is a non-measurable \(A\subseteq g(C)\).
The set \(B\coloneqq \inverseof{g}(A)\) is contained in \(C\), which has measure \(0\), so \(B\) is Lebesgue measurable and \(h\coloneqq\chi_B\) is a Lebesgue measurable function.
For \(y\in[0,2]\), \((h\circ\phi)(y) = 1\) if and only if \(\inverseof{g}(y)\in \inverseof{g}(A)\), that is, \(y\in A\). Hence \(\inverseof{(h\circ\phi)}(\theset{1}) = A\) is not Lebesgue measurable, and \(h\circ\phi\) is not Lebesgue measurable.
A homeomorphism can map a measurable set onto a non-measurable set (Tag PR-Z5VSQ)
There exist a homeomorphism \(\psi\colon[0,1]\to[0,2]\) and a Lebesgue measurable set \(Z\subseteq[0,1]\) such that \(\psi(Z)\) is not Lebesgue measurable.
Non-measurable sets
(The Vitali set)
Choose, using the axiom of choice, one representative in \([0,1]\) of each coset of \(\QQ\) in \(\RR\), and let \(V\) be the set of representatives. The translates \(V+r\) for \(r\in\QQ\cap[-1,1]\) are pairwise disjoint, their union contains \([0,1]\), and it is contained in \([-1,2]\). If \(V\) were Lebesgue measurable, translation invariance and countable additivity would give \(1\leq\sum_{r}m(V)\leq3\), which is impossible for \(m(V)=0\) and for \(m(V)>0\).
Every Lebesgue measurable subset of \(\RR\) of positive measure contains a non-measurable subset. The Lebesgue \(\sigma\)-algebra is the completion of the Borel \(\sigma\)-algebra and strictly contains it.
Exercises
The first Borel–Cantelli lemma (Tag P-QZT5B)
Show that if \(\sum \mu(E_k) < \infty\) then almost every \(x\in X\) is in at most finitely many \(E_k\).
Solution
Let \((X, \mu)\) be a measure space and \(E_k\) measurable. The set of points lying in infinitely many \(E_k\) is \(\limsup_k E_k = \bigcap_{n\ge1}\bigcup_{k\ge n} E_k\), which is measurable.
<1>1. \(\mu(\limsup_k E_k) \le \sum_{k\ge n}\mu(E_k)\) for every \(n\).
\(\limsup_k E_k \subseteq \bigcup_{k\ge n} E_k\), and countable subadditivity gives \(\mu(\bigcup_{k\ge n} E_k) \le \sum_{k\ge n}\mu(E_k)\).
<1>2. Q.E.D.
Since \(\sum_k \mu(E_k) < \infty\), the tails \(\sum_{k\ge n}\mu(E_k)\) tend to \(0\), so step <1>1 gives \(\mu(\limsup_k E_k) = 0\).
$E$ is measurable if and only if $E$ is regular (Tag P-TZJQI)
Show that \(E\) is measurable iff \(E\) is regular.
Solution
Setup note. For Lebesgue measure on \(\RR^n\), “regular” means: \(\mu^*(E) = \inf\{\mu(G) : G \supseteq E,\ G \text{ open}\}\) (outer) and \(\mu_*(E) = \sup\{\mu(F) : F \subseteq E,\ F \text{ compact}\}\) (inner, with \(\mu_*(E) = \mu^*(E)\)). We prove measurability is equivalent to regularity.
<1>1. Measurable \(\Rightarrow\) outer regular.
for measurable \(E\) with \(\mu(E) < \infty\), by definition of \(\mu^*\) there is a cover by boxes with total measure \(< \mu(E) + \eps\); the open set \(G\) obtained by enlarging the boxes slightly contains \(E\) and has \(\mu(G) \le \mu(E) + \eps\).
For \(\mu(E) = \infty\), exhaust by \(E \cap B(0,k)\) and take unions. Hence \(\mu(E) = \inf_{G \supseteq E \text{ open}}\mu(G)\). <1>2. Measurable \(\Rightarrow\) inner regular.
for measurable \(E\) with \(\mu(E) < \infty\), first approximate by closed sets: \(E^c\) is measurable, so by <1>1 there is open \(G \supseteq E^c\) with \(\mu(G \setminus E^c) < \eps\), giving closed \(F = G^c \subseteq E\) with \(\mu(E \setminus F) = \mu(G \setminus E^c) < \eps\).
Then approximate the closed set by compact sets (\(F \cap B(0,k) \nearrow F\)), and for infinite measure decompose \(E\) into finite-measure pieces. Hence \(\mu(E) = \sup_{F \subseteq E \text{ compact}}\mu(F)\). <1>3. Regular \(\Rightarrow\) measurable.
outer regularity gives open \(G_k \supseteq E\) with \(\mu(G_k) \le \mu^*(E) + 1/k\); let \(V = \cap_k G_k\), a \(G_\delta\) containing \(E\), with \(\mu(V) = \mu^*(E)\).
Similarly inner regularity gives compact \(F_k \subseteq E\) with \(\mu(F_k) \ge \mu_*(E) - 1/k\); let \(H = \cup_k F_k\), an \(F_\sigma\) inside \(E\), with \(\mu(H) = \mu_*(E)\). Regularity means \(\mu^*(E) = \mu_*(E)\), so \(E = V \setminus (V \setminus H)\) with \(V\) measurable and \(V \setminus H\) null (as \(\mu(V \setminus H) = \mu(V) - \mu(H) = 0\)); hence \(E\) is measurable. <1>4. Q.E.D.