Convergence in $L_p$ implies convergence in $L_1$

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I am trying to solve the following problem from “a course in probability theory “ by kai lai chung.

If $X_n \to X$ in $L_p$ and $Y_n \to Y$ in $L_q$ where $p>1$ and $\frac{1}{p}+\frac{1}{q} = 1$, then $X_nY_n \to XY$ in $L_1$.

My attempt: I can see that I have to somehow use holder’s inequality but can’t get the correct form where I can use it

$\mathbb{E}|X_nY_n-XY| = \mathbb{E}|X_nY_n -X_nY +X_nY -XY| \leq \mathbb{E}|X_n(Y_n -Y)| + \mathbb{E}|Y(X_n-X)|$

Now I could use holder for each term individually but that still leaves the single $X_n$ and $Y$.

So can someone please provide the solution.

Also does anybody know where I can find a solution manual or just solutions to kai lai chung?

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$$E|X_n Y_n - X Y| = E|X_n Y_n - X_n Y + X_n Y - X Y| \leq E|X_n(Y_n - Y)| + E|(X_n - X)Y| \leq (E|X_n|^p)^\frac{1}{p}(E|(Y_n - Y)|^q)^\frac{1}{q} + (E|Y|^q)^\frac{1}{q}(E|(X_n - X)|^p)^\frac{1}{p}$$

$(E|(Y_n - Y)|^q)^\frac{1}{q} \to 0$, $(E|(X_n - X)|^p)^\frac{1}{p} \to 0$ and $(E|X_n|^p)^\frac{1}{p} \to (E|X|^p)^\frac{1}{p}$ (and is thus bounded) because $X_n \to X$ in $L_p$ and $Y_n \to Y$ in $L_q$. Thus $E|X_n Y_n - X Y| \to 0$, Q.E.D.