Q:

Prove that $X_n \to X$ in $L^{2r}$ and $Y_n \to Y$ in $L^{2r}$ then $X_nY_n \to XY$ in $L^r$

My question concerns the following statement:

Suppose that $X, Y \in L^{2r}$ and $X_n \to X$ and $Y_n \to Y$ in $L^{2r}$. Show that $X_nY_n \to XY$ in $L^{r}$.
hint: $(a+b)^r \leqslant 2^{r-1}(a^r+b^r).$

I managed to get thus far:
$$E(|X_nY_n - XY|^r) = E(|X_n(Y_n - Y) + Y(X_n-X)|^r)$$
which is smaller equal than (triangle inequality + hint)

$2^{r-1} \left( E|X_n(Y_n-Y)|^r + E|Y(X_n-X)|^r\right)$

but I am stuck at this point. Anyone that can provide me with a hint?

A:

I believe that you almost did it. Using Cauchy's inequality we can get 
$$E(|Y(X_n-X)|^r)\leq E(|Y|^{2r})^{1/2}E(|X_n-X|^{2r})^{1/2}\rightarrow 0\ as \ n\rightarrow\infty.$$
As for the previous term, first use Cauchy's inequality and then notice that $E|X_n|^{2r}\rightarrow E|X|^{2r}$.
Addendum:$X_n\rightarrow X$ in $L^p$ implies that $E|X_n|^{p}\rightarrow E|X|^{p}$. Actually, Minkowski's inequality implies that $$(E|X|^p)^{1/p}=(E|(X-X_n)+X_n|^p)^{1/p}\leq (E|X-X_n|^p)^{1/p}+(E|X_n|^p)^{1/p}$$ and then we can conclude that $(E|X|^p)^{1/p} \leq \liminf_{n\rightarrow\infty}(E|X_n|^p)^{1/p}$ thus $E|X|^p \leq \liminf_{n\rightarrow\infty}E|X_n|^p$. The other direction can be shown in the same way. 

