Let $\{A_n\}_{n\in\mathbb{N}}$ be a sequence of disjoint measurable subsets of an interval $[a,b]$. Define the sequence of functions,
$$ f_n(x) = \frac{1}{\sqrt{m(A_n)}}\chi_{A_n}(x); \;\;\ x \in [a,b]$$
where $m$ is the Lebesgue measure. Prove or disprove:
1) For each $x \in [a,b], \lim\limits_{n\rightarrow\infty}f_n(x) = 0$
2) For any $g \in L^2[0,1], \lim\limits_{n\rightarrow\infty}\int_a^bf_n(x)g(x)dx = 0$
I believe 1) is true. If we fix an $x \in [a,b]$, then $x$ must belong to one and only one $A_n$, call it $A_N$.The for all $n \geq N$, $f_n(x) = 0$.
For 2), my initial idea was to try and dominate the integrand by some $L^1$ function to apply the DCT. But the denominator in $f_n$ is throwing things off. Is this statement even true?
By Cauchy-Schwarz, $$\left|\int_a^b f_n(x)g(x)dx\right| = \left|\frac{1}{\sqrt{m(A_n)}} \int_{A_n} g(x)dx\right|$$ $$ \le \frac{1}{\sqrt{m(A_n)}}\left(\int_{A_n} g(x)^2dx\right)^{1/2}\left(\int_{A_n} 1^2dx\right)^{1/2} = \left(\int_{A_n} g(x)^2dx\right)^{1/2} \to 0$$ since $g^2 \in L^1$ and $m(A_n) \to 0$.