How to prove convergence in $L^p$ imply convergence in $L^r$ when $p>r$? $X_n$ converges to $X$ in $p$th mean. Show that $X_n$ also converges to $X$ in $r$th mean when $p\ge r$.
I have tried conditioning on $|X_n-X|\ge1$ and $|X_n-X|<1$  but no luck. 
 A: As @user2566092 suggested, we have
$$\mathbb E\left[|X-X_n|^r\right]^{\frac pr} \leqslant \mathbb E\left[|X-X_n|^{\frac{rp}r}\right] = \mathbb E\left[|X-X_n|^p\right]. $$
Hence
$$\mathbb E\left[|X-X_n|^r\right] \leqslant \mathbb E\left[|X-X_n|^p\right]^{\frac rp}\stackrel{n\to\infty}\longrightarrow 0,$$
so that $X_n\stackrel{L^r}\longrightarrow X$. 
Note that this is only true in finite measure spaces (e.g. probability spaces). For example, consider $f:[1,\infty)$ with $f(x)=\frac1x$ and Lebesgue measure. Then
$$\int_{[1,\infty)}|f|\ \mathsf dm=\int_1^\infty\frac1x\ \mathsf dx = \lim_{x\to\infty}\log x = \infty, $$
so $f\notin L^1([1,\infty))$, but
$$\int_{[1,\infty)} |f|^2\ \mathsf dm = \int_1^\infty \frac1{x^2}\ \mathsf dx = 1 - \lim_{x\to\infty} \frac1x = 1<\infty, $$
so $f\in L^2([1,\infty))$.
A: Hint: Try to use Jensen's inequality with the convex function $f(Y_n)$ of the variable $Y_n = |X - X_n|^r$ defined by $f(Y_n) = {Y_n}^{p/r}$.
A: Indeed, you can bypass Jensen's inequality.
And as @Math1000 mentioned, suppose we are in a probability space. Then since $0<r\le p$ and by the linearity of expectation, we have:
\begin{align}
\mathbb E[|X_n-X|^r]&=\mathbb E[|X_n-X|^r\cdot 1_{|X_n-X|\ge 1}]+\mathbb E[|X_n-X|^r\cdot 1_{|X_n-X|<1}]\\
&\le\mathbb E[|X_n-X|^p\cdot 1_{|X_n-X|\ge 1}]+\mathbb E[|X_n-X|^r\cdot 1_{|X_n-X|<1}]\\
&\le\mathbb E[|X_n-X|^p]+\mathbb E[|X_n-X|^r\cdot 1_{|X_n-X|<1}]
\end{align}
Note that $\{|X_n-X|^p\}_{n=1}^\infty$ converging to $0$ in probability clearly implies that $|X_n-X|^r\cdot 1_{|X_n-X|<1}$ also converges to $0$ in probability(why?). And $||X_n-X|^r\cdot 1_{|X_n-X|<1}|<1$ uniformly together with $|X_n-X|^r\cdot 1_{|X_n-X|<1}$ converging to $0$ in probability imply that $|X_n-X|^r\cdot 1_{|X_n-X|<1}$ converges to $0$ in $L^1$(cf. Convergence in measure implies convergence in $L^p$ under the hypothesis of domination), i.e., $$\lim_{n\to\infty}\mathbb E[|X_n-X|^r\cdot 1_{|X_n-X|<1}]=0.$$
Thus,
$$ \lim_{n\to\infty}\mathbb E[|X_n-X|^r]\le\lim_{n\to\infty}\mathbb E[|X_n-X|^p]+\lim_{n\to\infty}\mathbb E[|X_n-X|^r\cdot 1_{|X_n-X|<1}]=0 $$
and we are done.
