Integral of the product of an bounded and any continous functions If $u:[0,1]\rightarrow \mathbb{R}$ is bounded measurable function so that for all $v\in C[0,1]$ 
$$\int_0^1uvdx=0$$then show that u is zero almost everywhere on $[0,1].$
Thanks in advance for any help!

Since $u$ is bounded and measurable, we have $u \in L^2[0,1]$. Since $L^2[0,1]$ is the completion of $C[0,1]$ with respect to the 'inner product' norm, we can find $u_n \in C[0,1]$ such that $u_n \to u$. Since $u_n$ is continuous, we have $\int u_n u  = 0$, and since $| \int (u-u_n)u | \le \|u-u_n\| \|u\|$, we see that
$\lim_n \int u_n u = \int u^2 = 0$. Hence $u(t) = 0$ ae. $t$.
If $u \in L^1[0,1]$ instead, we can apply a different technique. The function $\sigma(t) = 1_{[a,b]}(t)$ can be approximated (with the $\|\cdot\|_1$ norm) by uniformly bounded, continuous $\sigma_n$ so that $\sigma_n(t) \to \sigma(t)$ ae. $t$.
The dominated convergence theorem shows that $\lim_n \int u \sigma_n = \int u \sigma = \int_a^b u$. Since the $u_n$ are continuous, we have $\int_a^b u = 0$ for all $a,b \in [0,1]$. By considering the function $g(x) = \int_0^x u$, the Lebesgue differentiation theorem gives $g'(x) = u(x) $ ae. $x$, from which it follows that $u(x) = 0 $ ae. $x$.
