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Let $\mu$ be a measure on $(X,\mathcal A)$ and let $f, f_1, f_2,\dots$ and $g, g_1, g_2,\dots $be real valued $\mathcal A$- measureable functions on $X$.

Show that if $\mu$ is finite, $(f_n)$ convergence to $f$ in measure and $(g_n)$ to $g$ in measure, then $(f_ng_n)$ convergence to $fg$. Can the assumption that $\mu$ is finite be omitted?

Intuitively it feels like one could look at the measure of the union where the convergence does not hold, for $f_n$ and $g_n$. Is that correct or what should I do? And what about the finiteness of $\mu$?

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  • $\begingroup$ Maybe a better way to think of it. "Convergence in measure" may need an altered definition when $\mu$ is not finite. $\endgroup$ – GEdgar Aug 6 '13 at 14:49
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Finiteness of $\mu$ is needed: take $f_n(x):=\frac 1n\chi_{(0,n)}$ and $g(x)=g_n(x)=x$ where $\mu$ is the Lebesgue measure on the real line. Indeed, $f_n\to 0=:f$ in measure because for a fixed $\varepsilon$, for $n\gt 1/\varepsilon+1$, the set $\left\{x\in\mathbb R, \left\lvert f_n(x)\right\rvert\gt \varepsilon\right\}$ is empty. But it is not true that $f_ng_n\to 0$ in measure; for $\varepsilon=1/2$, $$ \left\{x\in\mathbb R, \left\lvert (f_ng_n)(x)\right\rvert\gt 1/2\right\} \supset \left(\frac n2,n\right).$$

When $\mu$ is finite, fix $\varepsilon>0$. There is $A>0$ such that $\mu\left\{|f|>A\right\}+\mu\left\{|g|>A\right\}<\varepsilon$.

Then write $$\left\{|f_ng_n-fg|>2\delta^2\right\}\subset\left\{|f_n-f|\cdot |g_n|>\delta^2\right\}\cup\left\{|f|\cdot|g_n-g|>\delta^2\right\}.$$ We can intersect over $\left\{|f|<A\right\}\cap \left\{|g|<A\right\}$ (its complement has a small measure). More precisely, let $B:=\left\{|f|<A\right\}\cap \left\{|g|<A\right\}$. Then $$\mu\left\{|f_ng_n-fg|>2\delta^2\right\}\leqslant \mu(B^c)+\mu(\left\{|f_n-f|\cdot |g_n|>\delta^2\right\}\cap B)+\mu(\{|f|\cdot|g_n-g|>\delta^2\}\cap B).$$ Since $\left\{|f|\cdot|g_n-g|>\delta^2\right\}\cap B\subset \left\{A\cdot |g_n-g|\gt\delta^2 \right\}\cap B\subset \left\{A\cdot |g_n-g|\gt\delta^2 \right\}$, the third term is not problematic. For the second one, notice that \begin{align} \mu\left(\left\{\left|f_n-f\right|\cdot \left|g_n\right|>2\delta^2\right\}\cap B\right) &\leqslant\mu\left\{|f_n-f|\cdot |g_n-g|>\delta^2\right\}+\mu\left(\left\{\left|f_n-f\right|\cdot |g|>\delta^2\right\}\cap B\right)\\ &\leqslant \mu\{\left|f_n-f\right|>\delta\}+\mu\{|g_n-g|>\delta\}+\mu\left\{\left|f_n-f\right|>\delta/A\right\}. \end{align}

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  • $\begingroup$ Can you please expand a little? and what do you mean with $\{\mu |f| > A \}$ I gues its $\mu \{x \in X : |f(x)| > A \}$? How do you get that inclusion with the $\delta$ and what does the intersection mean? $\endgroup$ – Johan Aug 6 '13 at 10:43
  • $\begingroup$ Yes, $\{|f|>A\}$ is the set of $x\in X$ for which $|f(x)|>A$. I've added details. $\endgroup$ – Davide Giraudo Aug 6 '13 at 11:04
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    $\begingroup$ @mezhang Because the sequences $\{|f|>n\}_{n=1}^\infty$ and $\{|g|>n\}_{n=1}^\infty$ are nested, and have $0$ measure. $\endgroup$ – Davide Giraudo Oct 16 '13 at 18:13
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    $\begingroup$ I meant "their intersection have $0$ measure". $\endgroup$ – Davide Giraudo Oct 16 '13 at 18:24
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    $\begingroup$ @TheHolyJoker I have added details on this part. $\endgroup$ – Davide Giraudo Dec 12 '19 at 11:00

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