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The following question is a qualifying exam question, though I don't see how these two parts are related.

(a)Let $(X, \mathcal F, \mu)$ be a measure space with $\mu(X)=1$ and suppose $F_1 , F_2, ...F_7$ are 7 measurable sets with $\mu(F_j) \geq 1/2$. Show that there exist indices $i_1<i_2<i_3<i_4$ for which $F_{i_{1}} \cap F_{i_{2}} \cap F_{i_{3}} \cap F_{i_{4}} \neq \varnothing $.

Thoughts: It seems this one is obvious I just suppose it's empty set then I can get a contradiction since the measure of the whole set is 1. I'm not sure whether it is the right track.

(b) Let $m$ denote Lebesgue measure on $[0,1]$ and let $f_n \in L^1 (m)$ be nonnegative and measurable with $\int_{[0,1/n]} f_n dm \geq 1/2$ for all $n \geq 1$. Show that $\int_{[0,1/n]} [\sup_{n} f_n] m(dx)=\infty$.

Thoughts: For this one, first I assume the integral of the sup is a finite number $M$ which is not infinity, then I can get the integral of the sup is going to zero,rather than $\geq 1/2$. I'm not sure about this idea either.

Any comments would be appreciated.

Edit: Thanks you guys. For (a), I'm wondering whether three sets also satifies the same result. I don't see why there always 4 such sets.

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  • $\begingroup$ For part (b), what about using $\int_{[0,1/n]} = \int_{[0,1/k]} + \int_{[1/k,1/n]}$? $\endgroup$ – Michael Sep 7 '15 at 4:30
  • $\begingroup$ Part (a) you are on the right track. The sum of the measures is at least 7/2 but there might be some "double counting." There is likely some way to use part (a) to get an alternative proof of (b), but I do not see it, I think my above comment would work just as well. $\endgroup$ – Michael Sep 7 '15 at 4:38
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For part (a): $$\int \sum_{j=1}^7\chi_{F_j}\ge 7/2>3.$$Since $\mu(X)=1$ there must exist $x$ with $\sum\chi_{F_j}(x)>3$.

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For Part (b), you can also do the following: Suppose that $g =\sup f_n$ is integrable. By the dominated convergence theorem, this yields $$ \int_{[0,1/n]} g \, dx \to 0, $$ but we have $$ \int_{[0,1/n]} g \, dx \geq \int_{[0,1/n]} f_n \, dx \geq 1/2, $$ a contradiction.

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