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Let's $A$ be such that $\lambda (A) = 0$ (Lebesgue measure). I want to prove that for every measurable function $f$,

$\int_A f(x) \lambda(dx) = 0$

I did the following : $|\int_A f(x) \lambda(dx)| <= \lambda(A) \times \sup_{x \in \mathbb{R}} |f(x)| = 0 \times \sup_{x \in \mathbb{R}} |f(x)| = 0$

For that to be true I have to assume that $0 \times \infty = 0$.

Is it an assumption we always do in measure theory, and why ? Or is there an other way to prove what I intended to prove ?

Bernard
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DimSum
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  • $\int_A f = \int f1_A$ and $f1_A = 0$ a.e. – mathworker21 Nov 11 '18 at 13:20
  • Thanks. So we never make assumption such that $0 \times \infty = 0$ in probability ? I have a vague souvenir where we did that in one of my probablity course. – DimSum Nov 11 '18 at 13:23
  • Secondly, this does not solve my problem how do you solve that if $g=0$ a.e. then $\int g = 0$, you would have to split the integral in two parts, one on set A where g=0 surely and the other part on the complementary of A which has measure 0. Thus we are back to $\int g = \int_{A^c} g = 0$ ? which is my initial problem.

    I know this is really basic stuff but this really bugs me.

     – DimSum Nov 11 '18 at 13:29
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    https://math.stackexchange.com/questions/293801/integral-over-null-set-is-zero – logo Nov 11 '18 at 13:34
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    from scratch: split $f$ into positive and negative parts, find a sequence of simple functions converging to $f$, and use the definition of the integral. – Matematleta Nov 11 '18 at 15:42
  • @DimitriMeunier yea you're right. I was hesitant to post that comment. But I don't like saying "we assume $0\times \infty = 0$ because that doesn't sound rigorous. Just appeal to the definition of the integral. You look at the sup of the integral over step functions that approximate. If a function is $0$ a.e., then any step function that is bounded above by it must be 0 identically. – mathworker21 Nov 11 '18 at 22:33

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