# If $f_n\to f$ in measure, then $f_ng\to fg$ in measure?

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.

• Is the measure finite? Do you assume $g$ to be integrable? Sep 29, 2021 at 16:53
• $g$ is not integrable. But the original question was to show $f_ng_n\to fg$. And the measure is finite. Sorry for missing that. Sep 29, 2021 at 16:55
• 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 $$\int\min(|f_ng-fg|,1)\,\mathrm d\mu=\int\min(|g||f_n-f|,1)\,\mathrm d\mu\ldots$$ Sep 29, 2021 at 16:57
• 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. Sep 29, 2021 at 17:12
• Right. In fact the assumption that $g$ be integrable is not needed. See my answer. Sep 29, 2021 at 17:19

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 $$A\to\infty$$.