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<a<b<\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.

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

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