proof that convergence in mean implies convergence in probability

Solution 1:

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 $$ \int_\Omega \mathbf{1}_{\{g_n>1\}}\,\mathrm dP\leq\int_\Omega g_n\,\mathrm dP=\frac{1}{\varepsilon^p}\int_{\Omega}|f_n-f|^p\,\mathrm dP. $$

Solution 2:

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 $(\lnot B \implies \lnot A)$.