Lebesgue measure: $\lim\limits_{n \to \infty} n \cdot m(S_n)=0$ where $S_n=\{x \in E \mid |f(x)|\geq n\} $

Question

Im looking to show that if $f$ is a Lebesgue integrable function on $E$ and if $$S_n=\{x \in E \mid |f(x)|\geq n\} $$ Then $$\lim_{n \to \infty} n \cdot m(S_n)=0 $$

We proved the Dominated convergence theorem, continuity from above, continuity from below, and a few other basic theorems from measure theory and Lebesgue theory. I'm not sure how to approach this type of problem since its not clear to be why this is important or how it relates to these topics.

Answer 1

Observe that $$\begin{aligned} \int_E |f| &= \int_{S_n} |f| + \int_{E \setminus S_n} |f| \\ &\geq \int_{S_n} n + \int_{E \setminus S_n} |f| \\ &= n\cdot m(S_n) + \int_{E \setminus S_n} |f| \\ \end{aligned}$$ As $\int_E |f|$ is finite, so is $\int_{E \setminus S_n} |f|$, so we can subtract the latter from both sides to obtain $$\begin{aligned} n\cdot m(S_n) &\leq \int_E |f| - \int_{E \setminus S_n} |f| \\ &= \int_{S_n} |f| \\ &= \int_E |f|\ \chi_{S_n} \\ \end{aligned}$$ where $\chi_{S_n}$ is the characteristic function of $S_n$. The integrand on the RHS is dominated by the integrable function $|f|$ and converges a.e. to zero, hence by the dominated convergence theorem we have $$\begin{aligned} \lim_{n \to \infty}n\cdot m(S_n) &\leq \lim_{n \to \infty}\int_E |f|\ \chi_{S_n} \\ &= \int_E \lim_{n \to \infty} |f|\ \chi_{S_n} \\ &= 0 \end{aligned}$$ As the LHS is nonnegative, this gives us the desired result.

Answer 2

Note that $n.m(S_n)=\int_{S_n}n\leq\int_{S_n}|f|$ for every $n$ and $\lim_{n\to+\infty}\int_{S_n}|f|=0$ because $f$ is integrable. So the proposition follows.