Proving an inequality for a Lebesgue integrable function

Question

Let $f:\mathbb{R}\to[-\infty , \infty]$ be integrable and $\epsilon > 0$. Show that

$m\{x:|f(x)|\geq \epsilon \} \leq \frac{1}{\epsilon} \int |f|$

I'm not sure how to do this one. My approach was to use the identity $|f|=f^++f^-$, where $f^+:=\max(f,0)$ and $f^-:=\max(-f,0)$. By assumption, $f$ is integrable so by definition $f$ is measurable and $\int f^+, \int f^- < \infty$. Then, as $f^+$ is non-negative, there is an increasing sequence of non-negative simple functions $(\phi_n)$ tending to $f^+$. Similarly a sequence $(\psi_n)$ tending to $f^-$. Then we can apply Monotone Convergence Theorem to get that

$\int |f| =\int (f^++f^-) = \lim_{n\to\infty} \int (\phi_n + \psi_n)$

But I'm not sure how to proceed. Any ideas?

Answer 1

This is called Chebyshev's inequality. Here is the proof:
Let $E := \{x:|f(x)| \geq \epsilon\}$ for convenience. Then
$m(E) \times \epsilon = \int_E \epsilon \ dm \leq \int_E|f| \ dm \leq \int|f| \ dm$
where the first inequality holds because $\epsilon \leq |f|$ on $E$.