Let $(X,\Sigma,\mu)$ be a finite measurable space ($\mu(X)<\infty$). Suppose $f_n \xrightarrow{\mu} f$ and $g_n \xrightarrow{\mu} f$, prove that $f_ng_n \xrightarrow{\mu} fg$

I'll write what I could do up to now:

Let $\lambda>0$, then $$\lambda<|f_ng_n(x)-fg(x)|$$$$=|f_ng_n-fg_n+fg_n-fg|$$$$\leq|g_n(x)||f_n(x)-f(x)|+|f(x)||g_n(x)-g(x)|$$

Let $S=\{x \in X: |f_ng_n(x)-fg(x)|>\lambda\}$, then $S \subset \{x \in X:|g_n(x)||f_n(x)-f(x)|>\dfrac{\lambda}{2}\} \cup \{x \in X:|f(x)||g_n(x)-g(x)|>\dfrac{\lambda}{2}\}$

If I call $S_1$ and $S_2$ to the first set and second sets of the union respectively, then given $N>0$, $$S_1 \subset A_1 \cup A_2,$$where $$A_1= \{|g_n| \geq N\dfrac{\sqrt{\lambda}}{4}\} \cap\{|f_n-f|<N^{-1}\dfrac{\sqrt{\lambda}}{4}\}$$ and$$A_2= \{|f_n-f| \geq N\dfrac{\sqrt{\lambda}}{4}\} \cap\{|g_n|<N^{-1}\dfrac{\sqrt{\lambda}}{4}\}$$

Similarly, $S_2 \subset B_1 \cup B_2$ with $$B_1= \{|f| \geq N\dfrac{\sqrt{\lambda}}{4}\} \cap\{|g_n-g|<N^{-1}\dfrac{\sqrt{\lambda}}{4}\}$$ and $$B_2=\{|g_n-g| \geq N\dfrac{\sqrt{\lambda}}{4}\} \cap\{|f|<N^{-1}\dfrac{\sqrt{\lambda}}{4}\}$$

I know how to find a bound for $A_2$ and $B_2$ using the fact that $f_n \xrightarrow{\mu} f$ and $g_n \xrightarrow{\mu} g$, but what about $A_1$ (or $B_1$)?

Thanks in advance


You work too hard. Suppose $f_n\rightarrow f$ and $g\rightarrow g$ in measure. Pass to a subsequence from $f_n$ so $f_n\rightarrow f$ a.e. Use the same subsequence and pass to a subsequence of $g_n$ so $g_n\rightarrow g$ and $f_n \rightarrow f$. Then $f_n g_n \rightarrow fg$ a.e. in this subsequence so $f_n g_n \rightarrow fg$ in $\mu$.

What have we shown? Every subsequence of $f_n g_n$ has a subsequence converging to $fg$ in measure. Since convergence in measure is a metric topology using metric $$ d(f,g) = \int_{\Omega} {|f-g|\over 1 + |f-g|}\, d\mu,$$ $f_ng_n \rightarrow fg$ in $\mu$.

  • 1
    $\begingroup$ Thanks for your answer but I haven't seen Lebesgue integral, I am supposed to solve this problem without it, so far we have covered measurable functions. $\endgroup$
    – user16924
    Apr 27 '15 at 0:00
  • $\begingroup$ Why is it sufficient for a subsequence to converge in measure? $\endgroup$
    – Keith
    Mar 5 '17 at 22:18
  • $\begingroup$ A sequence converges in measure iff every subsequence has a further subsequence converging a.s. $\endgroup$ May 3 '18 at 12:27

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