(Borel measurability) A function $f$ is said to be {\bf Borel measurable} provided its domain $E$ is a Borel set and for each $c$, the set $\{x \in E| f(x)>c\}$ is a Borel set. Show that
(i) if $f$ and $g$ are Borel measurable so are $af+bg$ and $fg$ where $a,b \in \mathbb{R}$.
(ii) every Borel measurable function is Lebesgue measurable.
(ii) if $f$ is Borel measurable and $B$ is a Borel set then $f^{-1}(B)$ is a Borel set.
(iii) if $f$ and $g$ are Borel measurable then $f \circ g$ is Borel measurable.
(iv) if $f$ is Borel measurable and $g$ is Lebesgue measurable then $f \circ g$ is Lebesgue measurable.
Proofs:
(i) If $a$ or $b$ is $0$, then this is trivial. If $a,b \in \mathbb{R}^{+}$, we may write
$$\{x: af+bg(x) > c\} = \bigcup_{\substack{s,r \in \mathbb{Q}^{+} \\ s\leq a, r \leq b }} \bigcup_{\substack{m,n \in \mathbb{Q} \\ m+n \leq c}} \{x: sf(x) > m \} \cap \{x : rg(x) > n \}, $$
which is measurable. Now, the other cases of $a,b \in \mathbb{R}^-$, $a \in \mathbb{R}^+, b \in \mathbb{R}^-$, and $a \in \mathbb{R}^-, b \in \mathbb{R}^+$ are handled similarly.
For $fg$, we may write \begin{align*} \{x : fg(x) > c\} &=
\left( \bigcup_{\substack{a,b \in \mathbb{Q}^+\\ ab \geq c}} \{x: f(x) > a \} \cap \{x : g(x) > b\} \right) \cup \left( \bigcup_{\substack{a \in \mathbb{Q}^-\\ b \in \mathbb{Q}^+ \\ ab \geq c}} \{x: f(x) < a \} \cap \{x : g(x) > b\} \right) \\
& \cup \left( \bigcup_{\substack{a \in \mathbb{Q}^+\\ b \in \mathbb{Q}^- \\ ab \geq c}} \{x: f(x) > a \} \cap \{x : g(x) < b\} \right) \cup \left( \bigcup_{\substack{a,b \in \mathbb{Q}^-\\ ab \geq c}} \{x: f(x) < a \} \cap \{x : g(x) < b\} \right)
\end{align*}
so $fg$ is measurable.
(ii) Every Borel Measurable set is Lebesgue Measurable, since if $B \in \mathcal{B}(\mathbb{R})$, then $B$ is the same as a Lebesgue Measurable set except possibly on a set of measure $0$.
(iii) Assuming $f: (X,\mathcal{T}) \to (\mathbb{R},\mathcal{U})$ with $(X,\mathcal{T})$ a general topological space, and $\mathcal{U}$ the standard topology on $\mathbb{R}$, by definition, any Borel set $B \in \mathcal{B}(\mathbb{R})$ is result of countable set operations on an open set. Now, given $f^{-1}((c,\infty)) \in \mathcal{B}(X)$, any open set may be written in terms of these open rays, and any Borel set in $\mathbb{R}$ may be written in terms of these open sets. Ergo, the inverse image of a Borel Set in $\mathbb{R}$ is the countable set theoretic result of operations on $f^{-1}((c,\infty))$, which is again a Borel Set, since $\mathcal{B}(X)$ is a $\sigma$-algebra.
(iii) Assume $g: \mathbb{R} \to \mathbb{R}$ and $f: \mathbb{R} \to \mathbb{R}$. Then, $(f \circ g)^{-1} ((c,\infty)) = g^{-1} \circ f^{-1} ((c,\infty))$. By hypothesis, $f^{-1} ((c,\infty)) = B\in \mathcal{B}(\mathbb{R})$. By definition of Borel Sets, any member of $\mathcal{B}(\mathbb{R})$ is the result of countable set operations on a member of the topology on $\mathbb{R}$. Any member of the topology on $\mathbb{R}$ may be written as the countable result of set operations on $(a,\infty)$ for some $a \in \mathbb{R}$, so $g^{-1} (B) \in \mathcal{B}(\mathbb{R})$, so $f \circ g$ is Borel Measurable.
(iv) I don't see what there is to prove here? Is it not the exact same argument as (iii) with a simple replacement of terminology?
Are the above proofs correct? What exactly is the purpose of (iv)?
Best Answer
It might be that you are not this far yet with your study, but I cannot withold myself from sharing this.
The Borelfunctions $f,g:E\rightarrow\mathbb R$ induce a function $h:E\rightarrow\mathbb R^2$ prescribed by $x\mapsto\langle f(x),g(x)\rangle$.
This $h$ is also a Borelfunction in the sense that $h^{-1}(B)\subseteq E$ is a Borelset for any $B$ that belongs to the Borel $\sigma$-algebra on $\mathbb R^2$.
Now if $p:\mathbb R^2\rightarrow\mathbb R$ is a Borelfunction then so is composition $p\circ h:E\rightarrow\mathbb R$.
Special case1: $p$ is prescribed by $\langle x,y\rangle\mapsto ax+by$.
Special case2: $p$ is prescribed by $\langle x,y\rangle\mapsto xy$.
In case1 $p\circ h$ can be recognized as $x\mapsto af(x)+bg(x)$
In case2 $p\circ h$ can be recognized as $x\mapsto f(x)g(x)$
So it solves case (i) on an elegant way.
Especially the existence of this easy route makes me reluctant to dive into your efforts.