proof that convergence in mean implies convergence in probability I'm attempting to understand a proof, but I am failing to see how a step is pulled off.
Claim: $\text{if } f_n \longrightarrow_{L_p} f$ then $f_n \longrightarrow_{P} f$
Proof: Let $\epsilon > 0$. Then
$P\left( \left\lbrace \omega: |f_n(\omega) - f(\omega)| > \epsilon \right\rbrace \right)  = P \left( \left\lbrace \omega: |f_n(\omega) - f(\omega)|^p > \epsilon^p \right\rbrace \right) \\$
$ \leq \frac{1}{\epsilon^p} \int_\Omega |f_n(\omega) - f(\omega)|^pdP(\omega) \longrightarrow_n 0$
I understand that this would show convergence in probability since by assumption the integral converges to zero. Its the step before that (moving from equality to inequality) that I am miffed by. This is my attempt:
$P \left( \left\lbrace \omega: |f_n(\omega) - f(\omega)|^p > \epsilon^p \right\rbrace \right) = \int_\Omega I[|f_n(\omega) - f(\omega)|^p > \epsilon^p]dP(\omega) \\ = \frac{\epsilon^p}{\epsilon^p} \int_\Omega I[|f_n(\omega) - f(\omega)|^p > \epsilon^p]dP(\omega) \\ = \frac{1}{\epsilon^p} \int_\Omega \epsilon^p I[|f_n(\omega) - f(\omega)|^p > \epsilon^p]dP(\omega) \\ \leq \frac{1}{\epsilon^p} \int_\Omega |f_n(\omega) - f(\omega)|^pdP(\omega) $ 
The last being from the fact that $\epsilon^p$ times the indicator will be $\epsilon^p$ or 0 unless the indicator is satisfied, which is less than $|f_n(\omega) - f(\omega)|$ for those $\omega$ since if the indicator is 1, $|f_n(\omega) - f(\omega)| > \epsilon^p$. But this feels way to loose (especially since we haven't changed the fact that were integrating over $\Omega$). Any help making this more clear/rigorous would be great.
Thanks!
 A: As an alternate, you can also use proof by contrapositive: we prove $(\lnot B \implies \lnot A)$ instead of $(A \implies B)$.
First, assume the negation of $B$, that is, $\{f_n\}$ does not converge in probability to $f$. Then there exists an $\varepsilon > 0$ for which
$$ \begin{align}
0 &< \lim_{n\to\infty} \mathbb{P}(|f_n-f| > \varepsilon) \\
& = \lim_{n\to\infty} \mathbb{P}(|f_n-f|^p > \varepsilon^p) \\
&\leq \lim_{n\to\infty} \frac{\mathbb{E}(|f_n-f|^p)}{\varepsilon^p} \quad (\text{Markov's Inequality})
\end{align}$$
So $\lim_{n\to\infty}   \mathbb{E}(|f_n-f|^p) \neq 0$, and then $\{f_n\}$ will not converge in $L^p$-norm to $f$. We hence proved $(\lnot B \implies \lnot A)$.
A: Since $\varepsilon>0$ we have
$$
|f_n-f|^p>\varepsilon^p\;\;\iff\;\;g_n:=\frac{|f_n-f|^p}{\varepsilon^p}>1,
$$
and hence
$$
\int_\Omega \mathbf{1}_{\{|f_n-f|^p>\varepsilon^p\}}\,\mathrm dP=\int_\Omega \mathbf{1}_{\{g_n>1\}}\,\mathrm dP.
$$
But $0\leq \mathbf{1}_{\{g_n>1\}}(\omega)\leq g_n(\omega)$ for all $\omega$ and hence
$$
\int_\Omega \mathbf{1}_{\{g_n>1\}}\,\mathrm dP\leq\int_\Omega g_n\,\mathrm dP=\frac{1}{\varepsilon^p}\int_{\Omega}|f_n-f|^p\,\mathrm dP.
$$
