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 $\mu$?

  • $\begingroup$ 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

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}

  • $\begingroup$ 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. $\endgroup$ Aug 6 '13 at 11:04
  • 1
    $\begingroup$ @mezhang Because the sequences $\{|f|>n\}_{n=1}^\infty$ and $\{|g|>n\}_{n=1}^\infty$ are nested, and have $0$ measure. $\endgroup$ Oct 16 '13 at 18:13
  • 1
    $\begingroup$ I meant "their intersection have $0$ measure". $\endgroup$ Oct 16 '13 at 18:24
  • 1
    $\begingroup$ @TheHolyJoker I have added details on this part. $\endgroup$ Dec 12 '19 at 11:00

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.