Let \((X,\mcm,\mu)\) be a measure space and let \(f\colon X\to\CC\) be measurable with \(\int_X \abs{f}\dmu = 0\). Then \(f = 0\) \(\mu\)-almost everywhere.
For each \(n \geq 1\) put \(E_n \coloneqq \theset{x\in X \suchthat \abs{f(x)} \geq \frac1n}\in\mcm\). On \(E_n\) we have \(\abs{f} \geq \frac1n\), so \[ 0 = \int_X \abs{f}\dmu \geq \int_{E_n} \abs{f}\dmu \geq \int_{E_n} \frac1n\dmu = \frac1n \mu(E_n), \] and hence \(\mu(E_n) = 0\) for every \(n\). Now \(\theset{x\in X \suchthat f(x) \neq 0} = \bigcup_{n \geq 1} E_n\) is a countable union of null sets, hence null. Therefore \(f = 0\) almost everywhere.