The $L^p$ norm on small sets is small

Question

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?

Answer 1

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.

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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.

Answer 2

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 ...