Integral over a set of measure 0

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 ?

• $\int_A f = \int f1_A$ 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. – Dimitri Meunier 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. – Dimitri Meunier Nov 11 '18 at 13:29
• math.stackexchange.com/questions/293801/… – logo Nov 11 '18 at 13:34
• 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