My definition for convergence in $L^p$ is the following;
A sequence $x_n \to x$ in $L^p$ to a random variable $x$ if $|x|^p$ is integrable and $E[|x_n -x|^p] \to 0$ as $n \to \infty$
My question is, if $x_n \to x$ in $L^p$ as $n \to \infty$ and $y_n \to y$ in $L^p$ as $n \to \infty$ then does $x_ny_n \to xy$ in $L^p$ as $n \to \infty$. Justify the answer by either providing a counter example or a proof.
I "think" its not true but i can't think of a counter example, could somebody please answer this.
Let the underlying probability space be the interval $[0,1]$ with Lebesgue measure as the probability measure.
Define $$f_a(t)=t^a$$ for $t\in(0,1]$. (It is undefined at $t=0$, which is no problem since $\{0\}$ has probability zero.) Now note the following:
First, note that this shows that the product of two $L^p$ variables do not need to belong to $L^p$: Just pick $a$ and $b$ so that $ap>-1$ and $bp>-1$, but $ap+bp \le -1$.
This should already give a strong hint that multiplication is not continuous even where it is defined. We work it out as follows:
Let $$a_n=\frac1p\Bigl(-\frac12+\frac1n\Bigr),\quad b_n=\frac1{n^{1/(2p)}},$$ and define $$x_n=b_nf_{a_n}.$$ You will find that $E[x_n^p]\to0$, i.e., $x_n\to0$ in $L^p$.
On the other hand, $x_n^2=b_n^2f_{2a_n}$, and so $$ E[(x_n^2)p]=\frac{b_n^p}{2a_np+1}=\frac12,$$ and so $x_n^2\in L^p$, but $x_n^2$ does not converge to zero in $L^p$.