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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.

Celine Harumi
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3 Answers3

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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.

zhw.
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  • Nice topological-like proof. Just a question. You showed that $f$ is Borel-measurable right? This implies it is Lebesgue measurable. – Celine Harumi Apr 29 '20 at 01:47
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    Yes that's right, $f$ is Borel measurable, hence Lebesgue measurable. – zhw. Apr 29 '20 at 03:38
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product of measurable functions is measurable, see here for a proof: Product of measurable functions is measurable

$g$ is continuous, therefore measurable, $I(\lvert x \rvert \leq 1)$ isa step function, so also measurable. Thus their product $f$, too.

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Suppose $A$ is measurable and $f:A \to \mathbb{R}$ is continuous. Then $L_\alpha=\{ x \in A | f(x) \le \alpha \}$ is closed relative to $A$ and hence measurable.

Define $\tilde{f}(x) = \begin{cases} f(x), & a \in A \\ 0, & \text{otherwise} \end{cases}$ (I will abuse notation slightly and write this as $\tilde{f} = f \cdot 1_A$.)

Then $\tilde{f}^{-1}((-\infty,\alpha])= \begin{cases} L_\alpha,& \alpha < 0 \\ L_\alpha \cup A^c, & \text{otherwise}\end{cases}$ and hence $\tilde{f}$ is measurable.

Returning to the question, suppose the points of discontinuities of $f$ are $x_1< x_2 < \cdots < x_n$ and let $I_0=(-\infty,x_1),I_1=(x_1,x_2)$, etc. $f$ is continuous on each of the $I_k$ and we have $f = f \cdot 1_{I_0} + f \cdot 1_{\{x_1\}} + f \cdot 1_{I_1} + \cdots$.

Since each of the components is measurable it follows that $f$ is measurable.

copper.hat
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