1. ## lebesgue measurable function

Q:
Prove that if $f : \mathbb{R} \longrightarrow [o,\infty]$ and $f^{-1}((r,\infty]) \in M$ for each $r \in \mathbb{Q}$, then $f$ is lebesgue measurable. ( $M$ is the $\sigma$-algebra of Lebesgue measurable sets)

Attempt:
I need to show that for any set $A \in \mathcal{B}_{[0,\infty]}$, $f^{-1}(A) \in M$ (is this correct?). So I take a set $A \in \mathcal{B}_{[0,\infty]}$. By definition of $\mathcal{B}_{[0,\infty]}$, $A$ is an open interval (is this correct?). Now, i want to write $A$ as a union of sets of the form $(r,\infty]$ for each $r \in \mathbb{Q}$ but im sturggling to do this! Is this the right method? Can you help me? Thank you

2. Hello,

I need to show that for any set $A \in \mathcal{B}_{[0,\infty]}, f^{-1}(A) \in M$ (is this correct?).
I think so, because the Lebesgue measure is a measure over the Borelian sigma-algebra of $\mathbb{R}$.

So I take a set $A \in \mathcal{B}_{[0,\infty]}$. By definition of $\mathcal{B}_{[0,\infty]}$, A is an open interval (is this correct?).
No, we just know that $\mathcal{B}_{[0,\infty)}$ is generated by the open intervals in $[0,\infty)$
In other words, it's generated by $\{(a,\infty) \mid a\in \mathbb{R}_+\}$
But an element of this will not necessarily be an open set.

It may be useful to use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$ (same goes for the positive numbers only).
Thus, $\forall a \in\mathbb{R}_+$, there exists a (decreasing) sequence $(a_n)_{n\geq 0}$, where $\forall n \in\mathbb{N} ~,~ a_n \in \mathbb{Q}_+$ such that $\lim_{n\uparrow \infty} a_n=a$
And we can write : $(a,\infty)=\bigcup_{n\geq 0} [a_n,\infty)$

But one can prove that $f^{-1}\left(\bigcup_{j\in J} A_j\right)=\bigcup_{j\in J} f^{-1}(A_j)$ (where J is a set without any restriction on it). It's one of Haussdorff formulas.

So you'll have $f^{-1}((a,\infty))\in M$, $\forall a\in \mathbb{R}_+$

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Now the second part (to finish) isn't easier imo.

There's a lemma that we called "Transport lemma", but I didn't find an English page of it... Here is the French one : Lemme de transport - Wikipédia
******transcription******
X,Y 2 sets. f : X -> Y. $\mathcal{E}\subset \mathcal{P}(Y)$ ( = the power set of Y = the family of all subsets of Y).
Then $f^{-1}(\sigma(\mathcal{E}))=\sigma(f^{-1}(\mathcal{E}))$

Here we have $\mathcal{E}=\{(a,\infty) \mid a\in\mathbb{R}_+\}$

But we know that $f^{-1}(\mathcal{E})\in M$
Hence, since M is a sigma-algebra, we have the (red) inclusion $f^{-1}(\sigma(\mathcal{E}))=\sigma(f^{-1}(\mathcal{E}))~{\color{red}\subset}~ M$

And thus for any element A in $\sigma(\mathcal{E})=\mathcal{B}_{[0,\infty)}$, $f^{-1}(A)\in M$

Whew... Does it look clear ? And I'm pretty sure there's a more elegant way to prove it, but I'm not an expert...
Well, you have a method, and maybe it will give you guidelines for building your own proof...
(as you can see, I'm very self-confident ^^)