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I am trying to solve the following Lebesgue integral problem:

Suppose $\mu(X)$ is finite. Let $f$ be an integrable function such that $f>0$ almost everywhere. For any $\varepsilon > 0$, there is a $\lambda>0$ such that $$\int_E f \,d\mu \geq \lambda$$ for all measurable $E$ with measure $\mu(E) \geq \varepsilon$

This question seems really intuitive in that set of nonnegative measure will have nonnegative integral, but I really have no idea where to start. Thank!

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    $\begingroup$ If you are on the whole line, then your result is definitely not true. Consider $f(x)=\frac{1}{x^2} \chi_{[1,\infty)}(x)$. Set $\varepsilon=1$ and consider $E_x=[x,x+1]$ as $x \to \infty$. In fact, if the measure space is infinite, then "$f$ is integrable" and the result you wish to prove are mutually exclusive. That is because the result you wish to prove lets you cut the space into infinitely many pieces, on each of which you have a fixed lower bound on the integral of $f$. $\endgroup$
    – Ian
    Oct 24, 2015 at 21:18

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Suppose $\mu$ is finite. Let $A_n = \{ x | f(x) \in [{1 \over n},0] \}$. Since $f(x) >0$ ae. [$\mu$], we see that $\mu A_n \to 0$. Choose $n$ such that $\mu A_n < { 1\over 2} \epsilon$.

Suppose $\mu E \ge \epsilon$, then $\mu (E \setminus A_n) \ge {1 \over 2 } \epsilon$ and $\int_E f d \mu \ge {1 \over n} {1 \over 2 } \epsilon$.

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  • $\begingroup$ Nice and easy, once you add the additional assumption to make it true in the first place. +1. :) $\endgroup$
    – Ian
    Oct 24, 2015 at 21:32
  • $\begingroup$ @Ian: Thanks! Practising my MathJax skills :-). $\endgroup$
    – copper.hat
    Oct 24, 2015 at 21:34
  • $\begingroup$ Actually, you might make explicit the use of continuity of measure in the third sentence. It is not so obvious in your exposition where you are actually using that $\mu$ is finite, and it is in that sentence. $\endgroup$
    – Ian
    Oct 24, 2015 at 23:53
  • $\begingroup$ @copper.hat It is a nice answer. If you allow me, I would suggest to two improvements: 1. when you write $[{1 \over n},0]$, I think you mean $[0,{1 \over n}]$; 2. After "Since $f(x) >0$ ae. [$\mu$]," you might include something like "we have that $A_n$ is a decresing sequence of sets and $\mu(\bigcap_{n}A_n)=\mu([f>0])=0$, so" and then continue with " we see that $\mu A_n \to 0$". $\endgroup$
    – Ramiro
    Oct 25, 2015 at 14:03

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