Prove the inverse of a lebesgue measurable function is measurable

Question

Okay so here is the question :

Let $I$ be an N-dimensional bounded interval and $f$ be a measurable (Lebesgue measurable) function on $I$. Show that

If $f \neq 0$ a.e. (almost everywhere) on $I$, then $\dfrac{1}{f}$ is also measurable on $I$.

So my thought process here is :

As $f \neq 0$ almost everywhere that means the set of points where $f=0$ is countable and has a measure zero. Hence, we have finite discontinuities if we consider $\dfrac{1}{f}$.

I'm pretty sure this won’t be enough. Anyone help me finish this? Or if wrong correct this?

Answer 1

You can easily check the following proposition.

Let $E\subset \mathbb R^N$ be a measurabe set. If $f$ and $g$ are two functions defined on $E$ satisfying $f=g$ for a.e.$x\in E$, then $f$ is measurable if and only if $g$ is measurable.

In this problem, we can thus assume $f\neq0$ on $I$. Since $\frac1x$ is a continuous function on $\mathbb R- \{0\}$, $\frac 1f$ is measurable.

Here we used this proposition:

$g \circ f$ is Lebesgue measurable, if $f$ is Lebesgue measurable and $g$ is continuous.

Answer 2

Just note that you need to prove that, for all $y\in \mathbb{R}$, the following set is a Borel set $$\left\{x\in \mathbb{R}:\frac{1}{f(x)}\le y\right\}=\left\{x\in \mathbb{R}:f(x)\ge 1/y\right\}\cup \left\{x\in \mathbb{R}:f(x)< 0\right\},$$ for $y>0$, and $$\left\{x\in \mathbb{R}:f(x)\le 1/y\right\}\cup \left\{x\in \mathbb{R}:f(x)\ge 0\right\},$$ for $y<0$. Let us write the shorthand $\{f\ge y\}$, to denote the above set. If $y>0$, then, as $f$ is Lebsegue measurable, the set is of course Borel. If $y\le 0$, the above sets are union of sets of the form $\{1/y\le f\}$(or its complement),$\ \{f\ge0\}$ (or its complement) , all of which are Borel sets (check!). Since $y$ was arbitrary, $1/f$ is Lebsegue measurable. In general, if $f$ is Lebesgue measurable, then for any function $g$ which is almost everywhere continuous, the function $g\circ f$ is measurable.