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This shouldn't be a hard problem, but I am stuck on it. I just need to prove the statement or come up with a counterexample. Any help will be appreciated.

Let $f: [0, 1] \rightarrow [0, \infty)$ be measurable (with respect to the restriction of Lebesgue measure to $[0, 1\textbf{]}$) with $\int_{[0, 1\textbf{]}}fdm = 1$. Does the following exist?

A measurable set $E$ and $n \in \mathbb{N}$ with $\frac{1}{n} \leq f \leq n$ on $E$ and $\int_Efdm \geq \frac{3}{4}.$

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    $\begingroup$ Just use the definition of integrable. You can always find a $M$, set $K = \{0\le f<M\}$ such that $\int_{E-K} f dm <\epsilon$ $\endgroup$ – Yimin Mar 7 '13 at 20:51
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Define $S_n:=\{x,\frac 1n\leqslant f(x)\leqslant n\}$ and $f_n:=f\chi_{S_n}$. What can we say about:

  • the pointwise limit of $f_n$?
  • the limit of $\int f_ndm$?
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