[B บ ....]

I want to show that the Haar functions in $L^2([0,1])$ forms an orthonormal basis :

Let $$f = 1_{[0, 1/2)} - 1_{[1/2,0)} \ \ \mbox{,} \ \ f_{j,k}(t) = 2^{j/2}f(2^jt - k).$$
Let $\mathscr{A} = \{(j.k) : j \geq 0, k = 0, 1, 2, ..., 2^j -1\}.$ I can prove that $\ A := \{1_{[0,1]}\} \cup \{f_{j,k}: (j,k) \in \mathscr{A}\}$ is an orthonormal system in $L^2([0,1])$.

(using the fact that each of them is supported on $[2^{-j}k, 2^{-j}(k+1))$, and each different pairs $i, j$ either has disjoint support or contained in each other support)

I want to show that $A$ is complete.

Let $g \in L^2([0,1])$ with $<g,f_{i,j}> = 0$ and $<g, 1_{[0,1]}> = 0$ for all $(i, j) \in A.$ I will show that $g = 0 $ a.e. Let $$I^l_{j,k} = [2^{-j},2^{-j}k + 2^{-j-1}), I^r_{j,k} = [2^{-j}k + 2^{-j-1}, 2^{-j}(k+1)).$$
Then $$f_{i,j} = 2^{-j}(1_{I^l_{i,j}} - 1_{I^r_{i,j}}).$$ So I see that $$\int_{I^l_{i,j}} f = \int _{I^r_{i,j}} f$$ for all $(i,j) \in A $ and $$\int_{[0.1]} f = 0.$$

It just "seems" that $f$ should be $0$ a.e., but how can I prove this rigorously ?

Any help please.