I am having trouble with the following practice question.  Hoping for a little help.  Thanks

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

(a) A measurable set $E$ and $n \in$ $\mathbb{N}$ with $\frac{1}{n} \leq f \leq n$ on $E$ and $\int_Ef dm \geq \frac{3}{4}.$

(b) A subinterval $J \subset [0, 1]$ of positive length and a subset $N \subset J$ of measure zero so that $f$ is bounded on $J\backslash N$.


For (a) define $E_n=\{ x\in [0,1]: \frac{1}{n}\leq f(x)\leq n\}$ and put $f_n= f\chi _{E_n}$ where $\chi_E$ is the characteristic function of the set $E$. Clearly $f_n\leq f $ and $f_n(x)\to f(x)$ everywhere. So $$ \int_{E_n} f \to \int_{[0,1]} f =1 $$ from which the answer is obvious.

For (b) take an enumeration of the rationals $q_j$ and define $$ g_j(x) = \frac{1}{\sqrt{|x-q_j|}} $$ $(g_j(q_j)=0)$. It's easy to see that $$ \int_{[0,1]} g_j(x)dx = \int_0^{q_j} \frac{1}{(q_j-x)^{1/2}}dx + \int_{q_j}^1 \frac{1}{(x-q_j)^{1/2}}dx \leq C $$ where $C$ is independent of $j$. Now define $$ g=\sum_{j=1}^{\infty} 2^{-j}g_j $$ By monotone convergence $g$ is integrable and unbounded in every subinterval in $[0,1]$ (and to make it bounded you would have to take away a neighbourhood of a rational point which is not of measure $0$).


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