Prove area under Lebesgue integrable function above $c$ can be made arbitrarily small with large enough $c$

Question

Let $f$ be a nonnegative Lebesgue integrable function on $[0,1]$. Denote by $m$ the Lebesgue measure on $[0,1]$. Prove that for reach $\epsilon > 0$, there is a $c >0$ such that $$\int_{\{ x \in [0,1] : f(x) \geq c\} } f dm < \epsilon.$$

Attempted Proof: By way of contradiction, letting $_c=\{x∈[0,1]:()\geq c\}$, for any $c$ we have that $$ \int_{[0,1] } f \geq \int_{E_c} f \geq c \cdot m (E_c) \geq c \cdot m (E)$$ where $E = \bigcap_{n=1}^\infty E_n$.

The problem is that $m(E)$ cannot be bounded below by a positive quantity, so I cannot make the integral arbitrarily large.

Answer 1

This is basically a version of Mitten's comment:

By possibly changing the values on a null set, we can assume that $f(x)$ is a real number for all $x\in[0,1]$. We have $$\int f~ \mathrm d m=\int_{[0,1]\setminus E_c} f~ \mathrm d m+\int_{E_c} f \mathrm d m.$$ Also, $\lim_{c\to\infty} 1_{[0,1]\setminus E_c} f(x)=f(x)$. By the dominated convergence theorem, $$\lim_{c\to\infty}\int_{[0,1]\setminus E_c} f~ \mathrm d m=\int f~ \mathrm dm.$$ So $$\lim_{c\to\infty}\int_{E_c} f~ \mathrm d m=\lim_{c\to\infty}\bigg(\int f~ \mathrm dm-\int_{[0,1]\setminus E_c} f~ \mathrm d m\bigg)=0.$$

Answer 2

I will write $\mu$ for the Lebesgue measure ( by way of habit) . Notice that as $f$ is positive and Lebesgue integrable the function $\nu: \mathcal{B}([0,1]) \rightarrow [0, \|f\|_{1} ] $ with:

$$ \nu(E) = \int_{E} f d\mu $$

is a finite measure absolutely continuous with respect to the Lebesgue measure. Furthermore notice that:

$$ \mu(E_{c}) \rightarrow 0 $$

This is a direct consequence of Markov's inequality , where $E_{c}$ as in your post, thus using a well known lemmma (actually a characterization of absolute continuity), we have that for some $\delta > 0 $ if $\mu(E_{c}) < \delta$ we must have:

$$ \nu(E_{c}) < \varepsilon$$

which is exactly what you want to prove.