# Prove the inverse of a lebesgue measurable function is measurable

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?

• BTW, the statement the set of points where $f=0$ is countable and has a measure zero in your post is not right. Sets of measurable zero can be uncountable, for example, the Cantor set. – Feng Shao Jul 14 '19 at 8:40
• Yeah thats the inverse of what I am saying. Countable sets have (Lebesgue) measure zero. – Feynstein Jul 14 '19 at 8:43

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.

• I see. I see 1/x is continuous, that proves $g\circ f$ is measurable. And if $g \circ f$ is measurable iff $g$ and $f$ are measurable. This is what you mean? Why did you mention the proposition for $g=f$? – Feynstein Jul 14 '19 at 8:45
• I worked on the setting of $\mathbb R$, so I mentioned the first proposition to make sure the composition of $\frac1x$ and $f$ makes sense. If you work on the extended real line, you can omit my first part. As the other answer mentioned in his last sentence. – Feng Shao Jul 14 '19 at 8:51
• BTW, the sentence "$g\circ f$ is measurable iff $g$ and $f$ are measurable" in your comment is not right. Watch out please. – Feng Shao Jul 14 '19 at 8:53
• Exactly, Then you have proved that $f \circ g$ is measurable. How do you propose to prove that $g$ is measurable? Where $g$ according to your answer can be taken as $g=1/x$ – Feynstein Jul 14 '19 at 8:54
• No,no. The symbols in these two propositions are independent. I apologize if it makes you unclear. Here is my though: Define $f_1$ such that $f_1(x)=f(x)$ if $f(x)\neq0$ and $f_1(x)=1\neq0$ if $f(x)=0$. From proposition 1, it suffices to prove the measurability of $\frac1{f_1}$, which is derived from proposition 2. In my answer I abused the notation to write $f_1$ as $f$. – Feng Shao Jul 14 '19 at 9:03

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.

• The identity $$\left\{x\in \mathbb{R}:\frac{1}{f(x)}\le y\right\}=\left\{x\in \mathbb{R}:f(x)\ge y\right\}$$ is not always right. – Feng Shao Jul 14 '19 at 7:27
• I guess you mean $$\left\{x\in \mathbb{R}:\frac{1}{f(x)}\le y\right\}=\left\{x\in \mathbb{R}:f(x)\ge \frac 1y\right\},$$ but this is not always right neither. – Feng Shao Jul 14 '19 at 7:29
• Oh, yes, sorry, I meant the latter. I think I also understand your point, which can be resolved as below: for $y>0$, $$\left\{x\in \mathbb{R}:\frac{1}{f(x)}\le y\right\}=\left\{x\in \mathbb{R}:f(x)\ge \frac 1y\right\}\cap \left\{x\in \mathbb{R}:f(x)\ge 0\right\},$$ and similar decomposition for $y<0$. – Samrat Mukhopadhyay Jul 14 '19 at 13:47
• I have added an explanation for resolving this issue in my answer. – Samrat Mukhopadhyay Jul 14 '19 at 13:53
• I know what you mean. But to be rigorous, for $y>0$,$$\left\{x\in \mathbb{R}:\frac{1}{f(x)}\le y\right\}=\left\{x\in \mathbb{R}:f(x)\ge 1/y\right\}\cup \left\{x\in \mathbb{R}:f(x)< 0\right\},$$ rather than $$\left\{x\in \mathbb{R}:\frac{1}{f(x)}\le y\right\}=\left\{x\in \mathbb{R}:f(x)\ge 1/y\right\}\cap \left\{x\in \mathbb{R}:f(x)\ge 0\right\}.$$ Am I right? – Feng Shao Jul 14 '19 at 13:56