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 $M=\int_{X\backslash E}|g|^pd\mu$.


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.

  • $\begingroup$ 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


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 $\epsilon\rightarrow0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.