Let $f:\mathbb{R}\to\mathbb{R}$ be bounded with compact support and has only finite points of discontinuities. Is $f$ Lebesgue measurable?
Details
In my case, $f$ has the form $f(x):=g(x)I(\lvert x \rvert \leq 1)$ with $g$ continuous. Hence, $f$ can be discontinuous at $x=1$ or $x=-1$. I want to show $\int_{-1}^1\lvert f(u) \rvert du<\infty$. For this, I will show that $f$ is bounded (which is easy) and that $f$ is Lebesgue measurable. If I show the measurability, then I can use the Lebesgue integral. From the boundedness
$$\int_{\mathbb{R}}\lvert f\rvert d\mu=\int_{-1}^1\lvert f\rvert d\mu=\int_{-1}^1\lvert f(u)\rvert du\leq M \int_{-1}^1du=2M<\infty,$$
for some $M>0$. The problem is that I don't know how to show that $f$ is measurable. I would appreciate any help or suggestion.
Best Answer
Suppose $f:\mathbb R\to \mathbb R$ is continuous on $U=\mathbb R\setminus F,$ where $F$ is a finite set. Note that $U$ is open.
Let $V\subset \mathbb R$ be open. Then
$$f^{-1}(V) = (f^{-1}(V)\cap U)\cup (f^{-1}(V)\cap F).$$
Because $f$ is continuous on $U,$ the first set on the right is open, hence measurable. The second set on the right is finite, hence measurable. The union of two measurable sets is measurable, and thus $f^{-1}(V)$ is measurable. It follows that $f$ is measurable.