Hello,

I think so, because the Lebesgue measure is a measure over the Borelian sigma-algebra of .I need to show that for any set (is this correct?).

No, we just know that is generated by the open intervals inSo I take a set . By definition of , A is an open interval (is this correct?).

In other words, it's generated by

But an element of this will not necessarily be an open set.

It may be useful to use the fact that is dense in (same goes for the positive numbers only).

Thus, , there exists a (decreasing) sequence , where such that

And we can write :

But one can prove that (where J is a set without any restriction on it). It's one of Haussdorff formulas.

So you'll have ,

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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

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X,Y 2 sets. f : X -> Y. ( = the power set of Y = the family of all subsets of Y).

Then

Here we have

But we know that

Hence, since M is a sigma-algebra, we have the (red) inclusion

And thus for any element A in ,

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 ^^)