Let $\mu$ be Lebesgue measure on $I=[0,1]$, and for $x\in I$, if $x=0.a_1a_2\ldots$ (written in binary), then let $R_n(x)=1$ if $a_n=1$, and $R_n(x)=-1$ if $a_n=0$. Let $S_n=\sum_{i=1}^n\left(\dfrac{1}{i}\right)R_i$. Show that $S_n$ converges in the $L^2$ sense to a function $H\in L^2$.
That function $H$ must be given by $\sum_{i=1}^{\infty}\left(\dfrac{1}{i}\right)R_i$. So I have to show that $H$ is in $L^2$, and $S_n$ converges in the $L_2$ sense to $H$.
But showing $H\in L^2$ means showing $\int_I H^2d\mu<\infty$. Since the terms that constitute the sum in $H$ can be positive or negative, I cannot use the monotone convergence theorem to evaluate this integral. How can I evaluate it?
Also, showing $S_n$ converges to $H$ means showing $\int_I (S_n-H)^2\rightarrow 0$ as $n\rightarrow\infty$. Again, monotone convergence theorem cannot be applied here.
The $R_n$ are the Rademacher functions. Note that $R_n(x)$ is not well-defined if $x$ is a rational number of the form $\frac{k}{2^n}$, then $x$ has two binary expansions, one that has only finitely non-zero bits, none of which is past the $n$-th place, and one that has only finitely many zero bits, none of which is past the $n$-th place. In one of the two representations, the $n$-th bit is $0$, in the other it is $1$. However, the set of all these rational numbers is a null set, and hence can be ignored for $L^2$ purposes.
We can most easily see that the Rademacher functions form an orthonormal family in $L^2([0,1])$ by extending them periodically to all of $\mathbb{R}$.
$$R_1(x) = \begin{cases}\;1 &, x - \lfloor x\rfloor \geqslant \frac12\\ -1 &, x-\lfloor x\rfloor < \frac12 \end{cases}$$
Then we have $R_k(x) = R_1(2^{k-1}x)$ for all $k$. Since $R_n(x)^2 \equiv 1$, we have $\lVert R_n\rVert_2 = 1$, and for $n < m$, each interval of constancy of $R_n$ consists of $2^{m-n-1}$ full periods of $R_m$, so the integral of $R_n(x)\cdot R_m(x)$ over such an interval is $0$. Since $[0,1)$ is the union of intervals of constancy of $R_n$, we have $$\int_0^1 R_n(x)R_m(x)\,dx = 0$$ for $n \neq m$.
Now the orthonormality quickly yields the result, for $n < m$, we have
$$\lVert S_m - S_n\rVert_2^2 = \left\lVert \sum_{k=n+1}^m \frac{1}{k}R_k\right\rVert_2^2 = \sum_{k=n+1}^m \frac{1}{k^2}\lVert R_k\rVert_2^2 = \sum_{k=n+1}^m \frac{1}{k^2} < \frac1n.$$
Thus $(S_n)$ is a Cauchy sequence, and by the completeness of $L^2([0,1])$, it converges to some $H \in L^2([0,1])$ (with $\lVert H\rVert_2^2 = \zeta(2) = \frac{\pi^2}{6}$).