The $L^p$ norm on small sets is small 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?
 A: A small point: monotone convergence doesn't apply here because $|f - f_m|^p$ does not increase as $m \to \infty$.
Perhaps you could define $$f_m = {\rm min}(f^+,m) - {\rm min}(|f^{-}|,m).$$
Since $f$ takes finite values almost everywhere (else its $L^p$ norm would be infinity), we know that $|f - f_m|^p$ converges pointwise to zero, everywhere except on the null set where the value of $f$ is infinite. Furthermore, $|f - f_m|^p$ is dominated by the integrable function $|f|^p$. So
$$ \lim_{m \to \infty} \int |f - f_m |^p d\mu = 0$$
by dominated convergence. Then you can apply the rest of your argument.

Just for fun, here is another argument that doesn't use Minkowski.
Suppose the contrary statement is true. Then for some $\epsilon > 0$, there exists a sequence of measurable subsets $A_n \subset S$ such that
$$ \mu(A_n) < \tfrac 1 {2^{n+1}}, \ \  \ \ \int |f|^p 1_{A_n} \geq \epsilon^p.$$
Define a new sequence $B_n = \cup_{m \geq n} A_m$. Then
$$ \mu(B_n) < \tfrac 1 {2^n}, \ \ \ \ \int |f|^p 1_{B_n} \geq \epsilon^p. $$
Finally, define $B = \cap_{n \geq 1} B_n$.
$B_n$ is a nested sequence of measurable sets, so the sequence of functions $|f|^p 1_{B_n}$ converges pointwise to $|f|^p 1_B$. The sequence is also dominated by the integrable function $|f|^p$, so by dominated convergence,
$$ \int |f|^p 1_B \geq \epsilon^p.$$
But the fact that $B_n$ is a nested sequence of sets of finite measure also implies that $$\mu(B) = \lim_{n \to \infty} \mu(B_n) = 0.$$
Hence $$ \int |f|^p 1_B = 0,$$
and we have a contradiction.
A: Hint for another approach: Simple functions are dense in every $L^p.$ The result we want is easy for simple functions, as they are bounded. So given $\epsilon>0,$ find a simple function $s$ such that $\|f-s\|_p < \epsilon$ and ...
