If $f$ is R-integrable, prove that if $\int_a^b |f(x)|dx = 0$, then $f(x) = 0$ for all $x\in[a,b]$. If this is not true, provide a counter-example.

Question

I've thought of multiple examples, none of which give $\int_a^b |f(x)|dx = 0$. So at this point I am thinking that a proof is in order. This is all that I have so far:

Let $F(x)=\int_{a}^{x}|f(t)|dt$. By assumption, $0=\int_{a}^{b}|f(t)|dt = \int_{a}^{x}|f(t)|dt+\int_{x}^{b}|f(t)|dt$ means, by the fundamental theorem of calculus, that $|F(x)-F(a)|+|F(x)-F(b)|=0 \Longrightarrow |F(x)-F(a)|=-|F(x)-F(b)|$.

This is the point where I start to get confused. Should I investigate cases or is there a simpler way of approaching this?

Answer 1

If$$f(x)=\begin{cases}1&\text{ if }x=a\\0&\text{ if }x>a,\end{cases}$$then $f$ is Riemann-integrable and $\int_a^b|f(x)|\,\mathrm dx=0$, but $f$ is not the null function.