If $\int f=0$, can I conclude that $f=0$?

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Suppose, $f:(0,1)\to\Re_+$ and $f\in L((0,1))$, therefore measurable. Is the following implication correct? $$\int_{(0,1)}f(x)\,d\mu=0\overset{?}{\implies} f(x)=0$$

Intuitively, I feel like the conclusion should hold because $f$ is a non negative function, but is there anyway of showing this?

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Take

$$f(x)=\begin{cases} &{1\over q}\quad x={p\over q}\in\Bbb{Q}\\ &0\quad x\in\Bbb{R}-\Bbb{Q} \end{cases}$$

This function is Lebesgue integrable with integral $0$