# The integral of a function is 0 in intervals, so the function is 0 everywhere

I am having trouble solving this problem of my course of measure and integration.

Let $$f$$ an integrable function on the measure space $$\mathbb{R}$$, $$L$$, $$\lambda$$, where $$L,\lambda$$ is the Lebesgue-measurable functions with Lebesgue measure. Such that: $$\displaystyle \int_a^b fd\lambda=0$$ for all $$-\infty. Prove that f=0 a.e.

I have tried to use the vanishing property, computing the integral over all $$\mathbb{R}$$, and distributing that like this: $$\displaystyle \int_\mathbb{R}fd\lambda=\int_{-\infty}^afd\lambda + \int_a^b fd\lambda +\int_b^{\infty} fd\lambda$$, note that the middle part equals to $$0$$. And trying to use the Countable Additivity of the Integral noticing that: $$(b,\infty) = \bigcup_{k=1}^{\infty}(b+k,b+k+1), \text{a.e.}$$ The problem is that I can't use the countable additivity becouse I have to asume that either $$f\geq0$$ or $$f \in L^{1}(b,\infty)$$. And I don't have any of that.

So if $$f<0$$, I wanted to divide in $$f=f_+ - f_-$$, where both are positive, but I don't know how to show that $$\int f= \int f_+ - \int f_-$$ becouse for doing that I have to asume that the last $$2$$ integrals exist and are not $$\infty$$. Any idea on how to do this?. Please help.

• See here – Brevan Ellefsen Mar 18 at 3:54
• Do you have Lebesgue's Differentiation Theorem? – Sean Haight Mar 18 at 3:57
• No, that exercise is after lebesgue dominated convergence teorema. And some other properties – J.Rodriguez Mar 18 at 14:11