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This was given on an old comp as a true or false problem:

If $1<p<\infty$, $|f_n|\leq 1$, $f_n\rightarrow f$ in measure, and $g_n\rightarrow g$ in $L_p$, then $f_ng_n\rightarrow fg$ in $L_p$.

I could not think of a counter example right away so here is my attempt:

Since convergence in $L_p$ implies convergence in measure $g_n\rightarrow g$ in measure. Let $E_1=\{x \in X:|f_n(x)-f(x)|>\epsilon_1\}$, $E_2=\{x \in X:|g_n(x)-g(x)|>\epsilon_2\}$ and $E=E_1\cup E_2$. Then $\mu(E)=0$ (warning this is incorrect usage of convergence in measure! I'll try to post a correct version in the answers)

and $$\int_X|f_ng_n-fg|^pd\mu=\int_{X\backslash E}|f_ng_n-fg+f_ng_n-f_ng_n|^pd\mu\leq\\\int_{X\backslash E}|f_n|^p|g_n-g|^pd\mu+\int_{X\backslash E}|g|^p|f_n-f|^pd\mu$$

The integral in the first part of the sum goes to zero but I'm not sure what to do with the second integral?

Is the fact that $g$ is in $L_p$ and $|f_n-f|^p<\epsilon_2^p$ enough? Since $$\int_{X\backslash E}|g|^p|f_n-f|^pd\mu\leq\epsilon_2^p\int_{X\backslash E}|g|^pd\mu=M\epsilon_2^p\rightarrow0$$ when $\epsilon_2\rightarrow0$ and $M=\int_{X\backslash E}|g|^pd\mu$.

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Notice that $$\lVert f_ng_n-fg\rVert_p\leqslant\lVert f_ng_n-f_ng\rVert_p+\lVert f_ng-fg\rVert_p $$ and using the fact that $|f_n|\leqslant 1$, we get $$\lVert f_ng_n-fg\rVert_p\leqslant\lVert g_n-g\rVert_p+\lVert (f_n-f)g\rVert_p,$$ hence the problem reduces to show that $\int|f_n-f|^p\cdot |g|^p\mathrm d\mu\to 0$. Now the idea suggested in the OP works: for a fixed $\varepsilon$, define $A_n:=\{|f_n-f|\gt\varepsilon\}$. Then $$\int|f_n-f|^p\cdot |g|^p\mathrm d\mu\leqslant \varepsilon^p+2^p\int_{A_n}|g|^p\mathrm d\mu.$$ Using an approximation argument, we can show that for each $\varepsilon$, $\int_{A_n}|g|^p\mathrm d\mu\to 0$, and the conclusion follows.

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  • $\begingroup$ Thank you for your answer. I do not see where the $2^p$ comes from. Would you mind elaborating that step for me? I just woke up so maybe I will see it in a bit. $\endgroup$ Aug 7 '14 at 14:33
  • $\begingroup$ I bound $|f_n-f|$ by $2$. $\endgroup$ Aug 7 '14 at 14:34
  • $\begingroup$ Ok got it. Just learning measure theory still getting used to all the $\frac{1}{2^n}$ type arguments that are used. $\endgroup$ Aug 7 '14 at 14:40
  • $\begingroup$ I think we can do that last step a little more directly using the dominated convergence theorem. By continuous mapping, $|f_n - f|^p |g|^p \to 0$ in measure, and is dominated by the integrable function $2^p |g|^p$. $\endgroup$ Aug 8 '14 at 16:16
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$$\int_X|f_ng_n-fg|^pd\mu=\int_X|f_ng_n-fg+f_ng_n-f_ng_n|^pd\mu\leq\\\int_X|f_n|^p|g_n-g|^pd\mu+\int_X|g|^p|f_n-f|^pd\mu$$

The first integral in the sum approaches $0$ as $n\rightarrow\infty$ since $g_n\rightarrow g$ in $L_p$ and $|f_n|\leq 1$.

For the second integral let $E_n:=\{x\in X:|f_n-f|>\frac{\epsilon}{2^n}\}$, $\int_X|g|^pd\mu=M$ (since $g\in L_p$), and note that $|f_n-f|\leq 2$ since $|f_n|\leq 1$ and some subsequence $f_{n_k}\rightarrow f$ almost uniformly. Thus

$$\int_X|g|^p|f_n-f|^pd\mu=\int_{X\backslash E_n}|g|^p|f_n-f|^pd\mu+\int_{E_n}|g|^p|f_n-f|^pd\mu\\\leq \frac{\epsilon^p M}{2^{np}}+2^pM\mu(E_n)$$

Which goes to $0$ as $n\rightarrow \infty$ and $\epsilon\rightarrow0$.

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