# Thread: lebesgue measurable function

1. ## lebesgue measurable function

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

Attempt:
I need to show that for any set $\displaystyle A \in \mathcal{B}_{[0,\infty]}$, $\displaystyle f^{-1}(A) \in M$ (is this correct?). So I take a set $\displaystyle A \in \mathcal{B}_{[0,\infty]}$. By definition of $\displaystyle \mathcal{B}_{[0,\infty]}$, $\displaystyle A$ is an open interval (is this correct?). Now, i want to write $\displaystyle A$ as a union of sets of the form $\displaystyle (r,\infty]$ for each $\displaystyle r \in \mathbb{Q}$ 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 $\displaystyle A \in \mathcal{B}_{[0,\infty]}, f^{-1}(A) \in M$ (is this correct?).
I think so, because the Lebesgue measure is a measure over the Borelian sigma-algebra of $\displaystyle \mathbb{R}$.

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

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

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

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

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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. $\displaystyle \mathcal{E}\subset \mathcal{P}(Y)$ ( = the power set of Y = the family of all subsets of Y).
Then $\displaystyle f^{-1}(\sigma(\mathcal{E}))=\sigma(f^{-1}(\mathcal{E}))$

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

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

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

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 ^^)