Show $\lim_{a \to 0} a \cdot \mu (\{ x \in \mathbb{R} : |f(x)| > a \}) = 0$ for $f \in L^1 (\mathbb{R})$, $a>0$

Question

Let $f \in L^1 (\mathbb{R})$, and $a > 0$. Show

$$ \lim_{a \to 0} a \cdot \mu (E_a) = 0 $$

where $E_a = \{ x \in \mathbb{R} : |f(x)| > a \}$.

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Try

Since

$$ \int_\mathbb{R} |f| d\mu = \int_{E_a} |f| d\mu + \int_{\mathbb{R}\setminus E_a } |f| d\mu < \infty $$

we have $\int_{E_a} |f| d\mu < \infty$, $\forall a$.

But I'm stuck at how I can relate this to find $\mu(E_a)$.

Any help will be appreciated.

Answer 1

It is a consequence of Lebesgue's dominated convergence theorem. Note that $$ a\cdot 1_{\{|f(x)|>a\}}\leq |f(x)| $$ for all $a>0$. Thus for any sequence $a_n$ such that $a_n \geq a_{n+1} \downarrow 0$, we have $$ \lim_{n\to\infty}a_n \cdot\mu(E_{a_n}) =\int \lim_{n\to\infty}\left(a_n\cdot 1_{\{|f(x)|>a_n\}}\right)d\mu =\int 0d\mu = 0. $$ This establishes $$\lim_{a\to 0} a\cdot\mu(E_a) =0.$$

Note: One can see that $a\cdot \mu(E_a)$ is the area of a rectangle below the graph $\{(x,|f(x)|) \;|\;x\in \mathbb{R}\}$ of $|f|$ which has finite area by the assumption. This observation is the motivation of this approach.