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Measure Theory Question

Matthew
Post Icon Posted: Submitted by Matthew on 12 April 2010 - 12:13am.

Joined: 2007-11-03
Posts: 12

From sheet 4 question A4b)
If $ f_n\rightarrow f $ in measure and $ g_n\rightarrow g $ in measure (for measureable $ f_n $ and $ g_n $) show that $ f_ng_n\rightarrow fg $ in measure if $ \mu(X)<\infty $, but not necessarily if $ \mu(X)=\infty $.
Here's what I've got:

Fix $ \epsilon>0 $ and $ \delta>0 $. Then there exists $ K>0 $ such that $ \mu(\{|f|>K\})<\epsilon $ and $ \mu(\{|g_n|>K\})<\epsilon\forall n $. Otherwise $ g_n $ and $ f $ are unbounded on a set with measure greater than 0. This is the bit I'm not sure about; does this give a contradiction with $ f $ and $ g_n $ being measurable or with $ g_n $ and $ f_n $ being Cauchy in measure? I figure it'd be pretty annoying taking infinity from infinity on a set of measure greater than zero when we look at $ \mu(\{x:|g_n(x)-g_m(x)|<\delta\}) $. Does it use $ \mu(X)<\infty $ (since the rest of the proof doesn't)? Or is this just wrong?

Anyway if this turns out to be true the rest of the proof follows like so:
\[|f_n(x)g_n(x)-f(x)g(x)|\leq|f_n(x)g_n(x)-f(x)g_n(x)|+|f(x)g_n(x)-f(x)g(x)|\]
\[=|g_n(x)||f_n(x)-f(x)|+|f(x)||g_n(x)-g(x)|\]
So
\[\mu(\{f_ng_n(x)-fg(x)|>\delta\})\leq\mu({x:|g_n(x)|>K\})+\mu(\{x:|f_n(x)-f(x)|>\delta/(2K)\}\]
\[+\mu({x:|f(x)|>K\})+\mu(\{x:|g_n(x)-g(x)|>\delta/(2K)\}\]
Take $ N $ such that $ n>N\Rightarrow $
\[\mu\left(\left\{x:|f_n(x)-f(x)|>\frac{\delta}{2K}\right\}\right)<\epsilon\]
\[\mu\left(\left\{x:|g_n(x)-g(x)|>\frac{\delta}{2K}\right\}\right)<\epsilon\]
So $ n>N\Rightarrow $
\[\mu(\{f_ng_n(x)-fg(x)|>\delta\})<4\epsilon\]

I can't see how to use $ \mu (X)<\infty $ or picture a counter example for $ \mu(X)=\infty $. Any ideas would be appreciated.
Cheers,
Matt

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