Lebesgue measure on $I=[0,1]$ Can you help me with this :)
$m$ is a Lebesgue measure on $I=[0,1]$, $g\in{L^{1}}(m)$ and 
$\int_{I} gf\, \mathrm{d}m=0$ for all $f\in{C(I)}$. Then I need to prove $g=0$ in $L^1(m)$?
 A: Choose a smooth approximation of unity (also known as mollifiers) $(\phi_\epsilon)_\epsilon$. Choose $x\in I$ and set $g=\phi_\epsilon(x-\cdot)$. Then
$$0=\int_I \phi_\epsilon (x-t) g(t) dt=(\phi_\epsilon*g)(x)\longrightarrow g(x)$$
as $\epsilon\rightarrow 0$ for a.e. $x$. Therefore $g(x)=0$ a.e.
A: If you want to bring this back to basics, you can use properties of outer Lebesgue measure, which is a common building block for defining Lebesgue measure. Using piecewise linear functions and Lebesgue bounded convergence, it's not hard to show that
$$
               \int_{a}^{b}f\,dx = 0,\;\;\; 0 \le a \le b \le 1.
$$
Every open subset of $\mathbb{R}$ can be written as a countable disjoint union of open intervals. So you'll be able to show that the following holds for every open subset $O\subset [a,b]$:
$$
                \int_{O}f\,dx = 0
$$
Building on outer measure: for a general measure subset $E$ of $(a,b)$, there is a sequence of open sets $\{ O_{j}\}_{j=1}^{n}$ such that (a) $E\subseteq O_{j}$ for all $j$, (b) $O_{j'} \subseteq O_{j}$ for all $j \le j'$, and (c) $\bigcap_{j}O_{j}\setminus E$ is a set of Lebesgue measure $0$. Therefore, by the dominated convergence theorem,
$$
          \int_{E}f\,dx = \int_{\bigcap_{j}O_{j}}f\,dx = \lim_{j}\int_{O_{j}}f\,dx =  0.
$$
Finally, for $\epsilon > 0$, Let $E_{\epsilon}=\{ x : f(x) \ge \epsilon \}$ in order to deduce that $mE_{\epsilon}=0$ from $0=\int_{E_{\epsilon}}f\,dx \ge \epsilon\cdot m E_{\epsilon} \ge 0$. This holds for all $\epsilon > 0$. Hence, $f \le 0$ a.e.. Repeat the argument for "$-f$" in order to conclude that $f=0$ a.e..
