I was stuck on the following question
If ${f_1, f_2, . . . }$ is a complete orthonormal set in $L^2 [0, 1]$ and $A$ is an arbitrary subset of positive Lebesgue measure in [$0,1]$ show that $$ \int_{A} \sum_{k=1}^{\infty} |f_{k}(x)|^{2} \geq 1$$
What I tried so far is to use to monotone convergence theorem to interchange the sum and the integral so what we are left with showing is $$ \sum_{k=1}^{\infty} \int_{A} |f_{k}(x)|^{2} \geq 1$$ I tried to assume the contrary that $$ \sum_{k=1}^{\infty} \int_{A} |f_{k}(x)|^{2} < 1$$ to that end I see that $f_{k}$ eventually vanishes on $A$ for large enough $k$, but I don't see how to get a contradiction.
You can prove it directly.
We have $1_A = \sum_k \langle f_k, 1_A \rangle f_k$ (in an $L^2$ sense) and Parseval combined with Cauchy Schwarz & Tonelli gives \begin{eqnarray} {\mu(A)} &=& \|1_A\|_2^2 \\ &=& \sum_k |\langle f_k, 1_A \rangle |^2 \\ &=& \sum_k |\langle f_k \cdot 1_A, 1_A \rangle |^2 \\ &\le & \sum_k \|1_A\|_2^2 \|f_k \cdot 1_A \|_2^2 \\ &=& \mu(A) \sum_k \int_A |f_k(x)|^2 dx \\ &=& \mu(A) \int_A \sum_k |f_k(x)|^2dx \end{eqnarray} Noting that $\mu(A) >0$ gives the desired result.