If $f$ is (Riemann) integrable on $[a,b]$ and if $\int_{a}^{b} fh=0$ for all continuous function $h$, then $f(x)=0$ for all points of continuity of $f$.
I know if we have $f$ being continuous on $[a,b]$, then we get $f(x)=0$ for all $x$ in $[a,b]$. But I have no idea about the case where $f$ is integrable. Can someone give me a hint? Thanks!
On
See this answer for a proof that for $f\in \mathscr R[a,b]$ and given any $\epsilon>0$, one can find $g\in\mathscr C[a,b]$ such that $$\int |f-g|<\epsilon. $$ Now note that $$|\int f^2|\le|\int (f^2-fg)|\le \int |f^2-fg|\le \sup |f|\int |f-g|\le \sup |f|\epsilon. $$ By our arbitrary choice of $\epsilon$ it follows that $$\int f^2=0. $$ Take it from here.
Hint:
Show that if $f$ is not $0$ at a continuity point, then $f(x) > 0$ or $f(x) < 0$ on some interval $[c_1,c_2]$.
Now choose
$$h(x) = \begin{cases} (x-c_1)(c_2-x), \,\, &x \in [c_1,c_2],\\ 0, \,\, &\mbox{otherwise}\end{cases}$$