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Let $f,g$ is a real-valued measurable function. Is it true that if $f_n\to f$ in measure, then $f_ng\to fg$ in measure?

The original question has the assumption that $g_n\to g$ in measure. And the question is to show $f_ng_n\to fg$ in measure. But I found that it boils down to show $f_ng\to fg$. I'm trying to use the analog of the proof in the case of the limit of functions, but the difficulty is that $g$ is not a constant, so $\frac{\epsilon}{g}$ is not fixed. I know this is duplicate, but I wonder if anyone could give me some hints. Thanks.

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  • $\begingroup$ Is the measure finite? Do you assume $g$ to be integrable? $\endgroup$
    – nejimban
    Sep 29, 2021 at 16:53
  • $\begingroup$ $g$ is not integrable. But the original question was to show $f_ng_n\to fg$. And the measure is finite. Sorry for missing that. $\endgroup$
    – jk001
    Sep 29, 2021 at 16:55
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    $\begingroup$ The convergence $f_n\to f$ in measure means that $$\int\min(|f_n-f|,1)\,\mathrm d\mu\xrightarrow[n\to\infty]{}0.$$ Now $$ \int\min(|f_ng-fg|,1)\,\mathrm d\mu=\int\min(|g||f_n-f|,1)\,\mathrm d\mu\ldots $$ $\endgroup$
    – nejimban
    Sep 29, 2021 at 16:57
  • $\begingroup$ Thanks so much. I certainly saw this equivalence relation before! I don't immediately see, and I feel like it would be easier if $g$ is integrable. $\endgroup$
    – jk001
    Sep 29, 2021 at 17:12
  • $\begingroup$ Right. In fact the assumption that $g$ be integrable is not needed. See my answer. $\endgroup$
    – nejimban
    Sep 29, 2021 at 17:19

1 Answer 1

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The convergence $f_n\to f$ in measure means that $$\int\min(|f_n-f|,1)\,\mathrm d\mu\xrightarrow[n\to\infty]{}0.$$ Now, for any $A>0$, \begin{align*} \int\min(|f_ng-fg|,1)\,\mathrm d\mu&=\int\min(|g||f_n-f|,1)\,\mathrm d\mu\\[.4em]&=\int_{\{|g|\le A\}}\min(|g||f_n-f|,1)\,\mathrm d\mu+\int_{\{|g|>A\}}\min(|g||f_n-f|,1)\,\mathrm d\mu\\[.4em]&\le A\int\min(|f_n-f|,1)\,\mathrm d\mu+\mu(\{|g|>A\}). \end{align*} Thus $$\limsup_{n\to\infty}\int\min(|f_ng-fg|,1)\,\mathrm d\mu\le\mu(\{|g|>A\}).$$ Assuming $g$ to be finite a.e., the right-hand side tends to $0$ as $A\to\infty$.

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  • $\begingroup$ Thanks! This is really nice... Just one question, can the first inequality be just the equality? $\endgroup$
    – jk001
    Sep 29, 2021 at 17:33
  • $\begingroup$ Yes it is an equality. I corrected it. $\endgroup$
    – nejimban
    Sep 29, 2021 at 17:41

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