Lebesgue integrable implies zero almost everywhere Let $f$ be an integrable function on $\mathbb{R}$ such that for every continuous $g$ on $\mathbb{R}$, $\displaystyle\int_\mathbb{R} fg \ dm=0$. Show that $f$ is zero a.e.
This is a qualifier sample problem and my idea is to proceed by contradiction. My problem here is that $f$ is not necessarily nonnegative so I'm having difficulties on establishing inequalities to arrive at a contradiction. Thanks!
 A: Hint: If $b\in L^\infty(\mathbb R),$ then there exists a sequence $g_n$ of bounded continuous functions on $\mathbb R$ such that $\|g_n\|_\infty\le \|b\|_\infty$ for all $n,$ and $g_n \to b$ pointwise a.e.
A: $\newcommand{\sign}{\operatorname{sign}}$By taking continuous functions $g$ with support in $(0,1)$, we can conclude that the statement is true for all functions $g\in C_c((0,1))$, i.e. functions with compact support in $(0,1)$, right? So let us focus on the function $f|_{(0,1)}$ on $(0,1)$ and forget about $\mathbb R$ for the moment. 
Since $(0,1)$ equipped with the Lebesgue measure $m$ is a finite measure space, Lusin tells us that for every $\varepsilon>0$ there exists a compact set $K_\varepsilon\subset (0,1)$ such that  $m((0,1)\setminus K_\varepsilon)<\varepsilon$ and a continuous function $h_\varepsilon$ with compact support such that $h_\varepsilon=\sign(f)$ on $K_\varepsilon$ and $\sup_{x\in (0,1)}|h_\varepsilon(x) |\leq 1$. This means for instance that we can take $g=h_\varepsilon$ and get
$$0=\int_{(0,1)}fg\,dm=\int_{K_\varepsilon} |f|\,dm + \int_{(0,1)\setminus K_\varepsilon}fh_{\varepsilon}\,dm$$
Now we take a sequence of $\varepsilon$s, say $\varepsilon_n=1/n$, and apply DCT (HOW?) to get that 
$$\int_{(0,1)}|f|\,dm=0$$
which implies $f=0$ a.e. on $(0,1)$. But we could also take $(a,b)$ instead of $(0,1)$, right? Of course, try it yourself and see if you understand it.
