Consider the following problem:

Let $E$ have finite measure, $\{f_n\} \rightarrow f$ in measure on $E$ and $g$ be a measurable function on $E$ that is finite almost everywhere. Use the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ to prove that $\{f_n^2 \} \rightarrow f^2$ in measure on $E$. Infer from this that if $\{g_n\} \rightarrow g$ in measure on $E$ then $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure on $E$.

I'm struggling understand how I can apply the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ in measure to prove that $\{f_n^2 \} \rightarrow f^2$ in measure. I've been able to come up with a proof of the fact that if $\{g_n\} \rightarrow g$ in measure then $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure, and this clearly implies the claim that $\{f_n^2 \} \rightarrow f^2$ in measure. Unfortunately my proof doesn't rely on the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ so it doesn't answer the problem since it explicitly asks us to prove it a certain way.

So far my best attempt to utilize the fact is as follows. Let $\eta > 0$ then by triangle inequality and monotonicity of measure

\begin{align*} \lim_{n\rightarrow \infty}\mu\left\{ \left|f_n^2 - f^2\right| > \eta \right\} &\leq \lim_{n\rightarrow \infty}\mu \left\{\left|f_n - f\right|\cdot \left|f_n\right| + \left|f_n - f\right| \cdot \left|f\right| > \eta \right\} \\ &\leq \lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right|\cdot \left|f_n\right| > \eta/2 \right\} + \lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right| \cdot \left|f\right| > \eta/2 \right\} \end{align*}

By the fact that $\{f_n \cdot f\} \rightarrow f \cdot f$ in measure \begin{equation*} \lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right| \cdot \left|f\right| > \eta/2 \right\} = \mu\left\{\left|f_n \cdot f - f\cdot f\right| > \eta/2 \right\} = 0. \end{equation*} I don't see how to deal with the other limit without first proving the most general statement $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure.

Any help or advice is greatly appreciated.


1 Answer 1


For fixed $M>0$ we clearly have

$$\begin{align*} \mu(|f_n-f| \cdot |f_n| > \eta/2) &\leq \mu(|f_n-f| \cdot |f_n|>\eta/2, |f_n| \leq M) \\ &\quad + \mu(|f_n-f| \cdot |f_n|>\eta/2, |f_n| > M), \end{align*}$$

and so

$$\mu(|f_n-f| \cdot |f_n| > \eta/2) \leq \mu(|f_n-f| >\eta/(2M)) + \mu(|f_n| > M). \tag{1}$$

As $f_n \to f$ in measure, the first term on the right-hand side converges to $0$ as $n \to \infty$. For the second one, we note that

$$\begin{align*} \mu(|f_n|>M) &= \mu(|f_n-f| < \eta, |f_n|>M) + \mu(|f_n-f|\geq \eta, |f_n|>M) \\ &\leq \mu(|f| \geq M-\eta) + \mu(|f_n-f| \geq \eta). \end{align*}$$

If we let $n \to \infty$, then $\mu(|f_n-f| \geq \eta) \to 0$ as $n \to \infty$. Hence,

$$\limsup_{n \to\infty} \mu(|f_n| >M) \leq \mu(|f| \geq M-\eta),$$

and therefore, by (1),

$$\limsup_{n \to \infty} \mu(|f_n-f| \cdot |f_n| > \eta/2) \leq \mu(|f| \geq M-\eta).$$

Since $\mu$ is a finite measure, we know that $\mu(|f| \geq M-\eta) \to 0$ as $M \to \infty$, and therefore we conclude that

$$\limsup_{n \to \infty} \mu(|f_n-f| \cdot |f_n| > \eta/2) = 0.$$

  • $\begingroup$ This is very nice thank you. I notice you spent a fair amount of time proving that $\lim_{M,n \rightarrow \infty} \mu(\left| f_n \right| > M) = 0$. In the Royden text (4th ed) it is assumed that by definition of convergence in measure, $f_n$ is finite a.e. Since this is the case the result seems to follow directly. Is the fact that $f_n$ is finite a.e. not generally part of the definition of convergence in measure? $\endgroup$
    – jodag
    Nov 5, 2018 at 22:15
  • 1
    $\begingroup$ @jodag Ah, no, it's not that simple; you have to be careful about taking the limits in the right order. In my proof, I first let $n \to \infty$ and afterwards (!) $M \to \infty$. I need to do it that way because otherwise I couldn't prove that the first term on the right-hand side of $(1)$ converges to $0$. This means that I need $$\lim_{M \to \infty} \lim_{n \to \infty} \mu(|f_n|>M)=0$$... which is not that obvious. (Your observation yields $\lim_{M \to \infty} \mu(|f_n|>M)$ for fixed $n \in \mathbb{N}$.) $\endgroup$
    – saz
    Nov 6, 2018 at 6:11
  • $\begingroup$ Ah I see. Thanks that helps. $\endgroup$
    – jodag
    Nov 6, 2018 at 20:59

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