Since the Haar system is a complete orthonormal system for $L^2 (\mathbb{R})$ any $L$-square function $f$ can be written as a superposition of elements of the Haar system: $$ f(x) = \sum_j \sum_k \langle f,h_{j,k} \rangle h_{j,k}(x) $$ where $$h_{j,k}(x) = \begin{cases} 1 & x\in [\frac{k}{2^{j}},\frac{k+1/2}{2^{j}} )\\ -1 & x\in [\frac{k+1/2}{2^{j}},\frac{k+1}{2^{j}} ) \\ 0 & otherwise \end{cases}$$
Integration yields
$$ \int_{\mathbb{R}} f(x)dx = \int_{\mathbb{R}} \sum_j \sum_k \langle f,h_{j,k} \rangle h_{j,k}(x)dx = \sum_j \sum_k \langle f,h_{j,k} \rangle \int_{\mathbb{R}} h_{j,k}(x)dx = 0$$ which is to say any any $L$-square function $f$ has zero integral over the real line which is pure nonsense. (We might as well take $f$ to be compactly supported but that doesn't change anything.) What have I done wrong? Thanks.
There are two points. First, not every function in $L^2(\mathbb{R})$ is integrable. And second, on the subspace $L^1(\mathbb{R}) \cap L^2(\mathbb{R})$, the integral is not continuous with respect to the $L^2$-norm. Since the series
$$f = \sum_{j,k} \langle f, h_{j,k}\rangle h_{j,k}$$
converges to $f$ in the $L^2$-norm, but usually not in the $L^1$-norm (if $f \in L^1\cap L^2$), one can in general not move the sum outside the integral.