A function that is measurable on $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$

Question

This seems to be a result that is used often, which I am having difficulty proving. There may be a more general statement, but I'll write out the statement that I was trying to prove:

Let $f:(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n)) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ be a function that is continuous on a set $A \in \mathcal{B}(\mathbb{R}^n)$, which may not necessarily be continuous on $A^c$. Then, $f \chi_{A}$ is measurable in $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$, where $\chi$ is the indicator function.

My attempt was to start on the set $A$ and artificially 'extend' $f \chi_{A}$ continuously to all of $\mathbb{R}^n$. But I met many difficulties, as the set $A$ may not be convex, and in general borel sets are hard to visualize. Does this have a neat solution?

Answer 1

A direct approach:

EDITED ANSWER

Since $f\chi_A(x)=\begin{cases}f(x)&x\in A\\0&x\notin A\end{cases}$, take $a\in\Bbb R$.

If $a> 0$, then $$\{f\chi_A(x)< a\}=\{x\in A:\, f(x)< a\}\cup A^c$$ Since $f$ is continuous on $A$, the first set is a (relative) open set on $A$, so measurable ($A$ is measurable). So both sets are measurables and we are done.

Now, if $a\le 0$, then $$\{f\chi_A(x)< a\}=A\cap\{f(x)<a\}$$ and again is measurable using the same argument above.

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PREVIOUS (not correct) ANSWER

If $a\ge 0$, then $$\{f\chi_A(x)\le a\}=\{x\in A:\, f(x)\le a\}\cup A^c=(A\cap\{f(x)\le a\})\cup A^c$$ and this set is measurable since both $A$ and $f$ so are.

Now, if $a<0$, then $$\{f\chi_A(x)\le a\}=A\cap\{f(x)\le a\}$$ and again is measurable since $A$ and $f$ are.

Answer 2

Let $A$ be a set which is not Borel. Let $f=\chi_A$. Then $f$ is continuous on $A$ (w.r.t the realtive topology from $\mathbb R^{n}$ or any other topology!). But $f\chi_A=\chi_A$ is not Borel measurable since the inverse image of $\{1\}$ is not Borel set. Hence, neither $f$ nor $f\chi_A$ has to be measruable.

However, if $A$ is a Borel set then $f$ and $f\chi_A$ are Borel meaurable: For any open set $U$ in the real line $f^{-1}(U)$ is an open set in $A$ which means there is an open set $V$ in $\mathbb R^{n}$ such that $f^{-1}(U)=A\cap V$. It folows that $f^{-1}(U)$ is Borel.