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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?

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  • $\begingroup$ 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. $\endgroup$ – Feng Shao Jul 14 '19 at 8:40
  • $\begingroup$ Yeah thats the inverse of what I am saying. Countable sets have (Lebesgue) measure zero. $\endgroup$ – Feynstein Jul 14 '19 at 8:43
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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.

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  • $\begingroup$ 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 $g=f$? $\endgroup$ – Feynstein Jul 14 '19 at 8:45
  • $\begingroup$ 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. $\endgroup$ – Feng Shao Jul 14 '19 at 8:51
  • $\begingroup$ BTW, the sentence "$g\circ f$ is measurable iff $g$ and $f$ are measurable" in your comment is not right. Watch out please. $\endgroup$ – Feng Shao Jul 14 '19 at 8:53
  • $\begingroup$ 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 $g=1/x$ $\endgroup$ – Feynstein Jul 14 '19 at 8:54
  • $\begingroup$ 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 $f$. $\endgroup$ – Feng Shao Jul 14 '19 at 9:03
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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.

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  • $\begingroup$ 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. $\endgroup$ – Feng Shao Jul 14 '19 at 7:27
  • $\begingroup$ 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. $\endgroup$ – Feng Shao Jul 14 '19 at 7:29
  • $\begingroup$ 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 $y<0$. $\endgroup$ – Samrat Mukhopadhyay Jul 14 '19 at 13:47
  • $\begingroup$ I have added an explanation for resolving this issue in my answer. $\endgroup$ – Samrat Mukhopadhyay Jul 14 '19 at 13:53
  • $\begingroup$ 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? $\endgroup$ – Feng Shao Jul 14 '19 at 13:56

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