$f,g$ are lebesgue-measurable and finite a.e. Then, prove that $f\cdot g$ is lebesgue-measurable.

Question

Let $f,g : \mathbb{R^n} \to [-\infty, \infty]$ are lebesgue-measurable functions and $f,g$ are finite a.e.

Then, prove that $f\cdot g$ is lebesgue-measurable by using the fact that $h(x,y)=xy$ defined on $\mathbb{R^2}$ is continuous.

My attempt is here.

Let $F:= h(f,g)=f\cdot g$. Then, it is enough to prove that $F$ is lebesgue-measurable.

Since $h$ is continuous, $F$ is also continuous.

Let $N_1:= \{ x \in \mathbb{R^n} \ | \ f(x)=\pm \infty \},$ and $N_2=\{ x \in \mathbb{R^n} \ | \ g(x)=\pm \infty \}$. Since $f$ and $g$ are finite a.e, $m(N_1)=m(N_2)=0.$

For all $a\in \mathbb{R},$

\begin{align} \{ F > a \} &=\Bigg( \{ F > a \}\cap (N_1 \cup N_2)\Bigg) \bigcup \Bigg(\{ F > a \}\cap (N_1 \cup N_2)^c\Bigg) \end{align}

Let $A:=\{ F > a \}\cap (N_1 \cup N_2)$ and $B:=\{ F > a \}\cap (N_1 \cup N_2)^c$.

Since $A \subset N_1\cup N_2$, $m(A)\leqq m(N_1)+m(N_2)=0$ thus $m(A)=0$ and $A$ is lebesgue-measurable.

$B=\{ x \in \mathbb{R^n} \ | \ F(x)> a, F(x)\neq \pm \infty \} =F^{-1}((a, \infty))$

Since $F$ is continuous, $F^{-1}((a, \infty))$ is open and therefore lebesgue-measurable.

Thus, $\{ F > a \} =A\cup B$ is lebesgue-measurable.

I wonder if this proof is correct.

At the beginning, I let $F:=h(f,g)$. But is this possible? $h$ is defined on $\mathbb{R^2}$, but $f,g$ can be $\pm \infty$. So if $f=\infty$ and $g=\infty,$ $F=h(\infty, \infty)$. This cannot be defined.

Answer 1

Well, things go wrong early on here.

Let $F:= h(f,g)=f\cdot g$. Then, it is enough to prove that $F$ is lebesgue-measurable.

Since $h$ is continuous, $F$ is also continuous.

This statement is not true. Consider $f(x) = \begin{cases} 0 & x \le 0 \\ 1 & x > 0\end{cases},$ and let $g = f$. Then $F(x) = h(f(x), f(x)) = f(x) \cdot f(x) = f(x),$ which is discontinuous.

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BTW, you don't appear to use the continuity of $F$ anywhere else in the proof, so just deleting the sentence about the continuity of $F$ might be a good first step.

Answer 2

Composition rule: If $u:(\mathrm{X}, \mathscr{X}) \to (\mathrm{Y}, \mathscr{Y})$ and $v:(\mathrm{Y}, \mathscr{Y}) \to (\mathrm{Z}, \mathscr{Z})$ are measurable, then $v \circ u:(\mathrm{X}, \mathscr{X}) \to (\mathrm{Z}, \mathscr{Z})$ is also measurable.

If $f$ and $g$ were Borel-measurable, the claim would be obvious by the composition rule above, since then $(f,g)$ would be Borel measurable from $\mathbf{R}^n \to \mathbf{R}^2$ (Borel $\sigma$-algebras show that the measurable space $\mathbf{R}^2$ coincides with the product space $\mathbf{R} \times \mathbf{R},$ this is not true for Lebesgue $\sigma$-algebras). In other words, the product of two Borel measurable functions is Borel measurable.

Consider $f$ and $g$ two be two measurable functions $\mathbf{R}^n \to \mathbf{R},$ then there exists Borel measurable functions $f_0$ and $g_0$ such that $f = f_0$ a.e. and $g = g_0$ a.e. A fortiori, there exists a null set $\mathrm{N}$ such that $f = f_0$ and $g = g_0$ on $\mathbf{R}^n \setminus \mathrm{N}.$ Write $fg = f_0g_0 + (f - f_0)(g - g_0) \mathbf{1}_\mathrm{N}.$ We know $f_0g_0$ is Borel measurable and since $(f - f_0)(g - g_0) \mathbf{1}_\mathrm{N}$ is zero a.e., it is measurable, a fortiori $fg$ is measurable. Q.E.D.