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I will write a question from Folland's book. What I want to ask is not the solution of this problem, but the way how to approach it. Question is as follows:

If $f \in L^+$ and $\int f < \infty$, for every $\epsilon > 0$ there exists $E \in \mathcal{M}$ such that $\mu(E) < \infty$ and $\int_E f > (\int f) - \epsilon$.

So as I said, I simply need to understand the approach I should take. For instance, what does the last inequality mean? What it says when you write a statement like $b > a - \epsilon$?

Also, I want to gain intuition about the way of full solution. Thanks.

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What about the measurable space we are working one (I guess the total mass is infinite in order to avoid a trivial problem)? Can you do it when $f$ is the characteristic function of a set of finite measure? What the inequality gives is $\int_{X\setminus E}f<\varepsilon$, that is, the integral of $f$ is "almost" concentrated on a set of finite measure. – Davide Giraudo Nov 7 '12 at 22:12
What does L+ stand for? – Amr Nov 7 '12 at 22:13
@Amr: Fix a measure space $(X, \mathcal{M}, \mu)$, then $L^+$ is the space of all measurable functions from $X \to [0,\infty]$. – oeda Nov 7 '12 at 22:15
I can only prove the theorem for sigma finite measure spaces – Amr Nov 7 '12 at 22:20
@Amr: the support of an integrable function is necessarily $\sigma$-finite (in fact, integrable functions form a dense subset of the set of all $\sigma$-finitely supported functions ), so assuming that $\mu$ is $\sigma$-finite is not much of a restriction. – tomasz Nov 7 '12 at 22:51
up vote 2 down vote accepted

We have to show that for each $\varepsilon>0$, we can find $S_{\varepsilon}$ which concentrates the integral up to $\varepsilon$, i.e. $$\int_{X\setminus S_{\varepsilon}}fd\mu<\varepsilon$$ and is of finite measure. We can try, and expand, the following sketch of proof:

  • Using the definition of Lebesgue integral, we can restrict ourselves to simple functions.
  • By linearity, we can do a further restriction to characteristic function of measurable sets of finite measure.
  • This case is easy.

Note that this doesn't need $\sigma$-finite of the measure space.

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