Let $(S,\Sigma,\mu)$ be a measure space. Let $f \in L^p(S)$. Let $\epsilon >0$. There exists $\delta >0$ such that if $A \in \Sigma$ with $\mu(A) < \delta$, then 

$$\|f\mathbb{1}\_A\|\_{L^p} < \epsilon.$$

Let $f\_m(x) = \min(f(x),m)$. Then $f\_m \uparrow f$ pointwise and we may apply monotone convergence. In particular, let $M$ be such that $$\int\_{S} |f-f\_M|^p \, d\mu < \left( \frac{\epsilon}{2} \right)^p$$ then, let $\delta < \left( \frac{\epsilon}{2M} \right)^p$. Choose $A \in \Sigma$ such that $\mu(A) < \delta$. Then by Minkowski,
\begin{align\*}
\|f\mathbb{1}\_A\|\_{L^p} &\leq \|(f-f\_M)\mathbb{1}\_A\|\_{L^p} + \|f\_M\mathbb{1}\_A\|\_{L^p} \\
&< \frac{\epsilon}{2} + M\frac{\epsilon}{2M} = \epsilon
\end{align\*}

Is this a valid proof?