Sum and product of functions convergence in measure

**tttcomrader** · November 14th 2009, 07:35 PM

Suppose $f_n \rightarrow f$ in measure, and $g_n \rightarrow g$ in measure, prove that (a) $f_n+g_n \rightarrow f+g$ in measure and that (b) $f_ng_n \rightarrow fg$ in measure if $\mu (X) < \infty$ , not if $\mu (X) = \infty$

Proof so far.

(a) I think this is pretty easy.

I know that $\forall \epsilon > 0$ , we have $\mu ( \{ x : \mid f_n(x)-f(x) \mid \geq \epsilon ) \rightarrow 0$ and $\mu ( \{ x : \mid g_n(x)-g(x) \mid \geq \epsilon ) \rightarrow 0$ .

Now, since $\{ x: \mid f_n(x)+g_n(x)-f(x)-g(x) \mid \geq \epsilon \}$ $\subset \{ x: \mid f_n(x)-f(x) \mid \geq \frac { \epsilon }{2} \} \cup \{ x: \mid g_n(x)-g(x) \mid \geq \frac { \epsilon }{2} \}$

Question 1: Is this really right? I took this from a proof in the book, but I'm failing to fully understand why it is true.

So it follows that $\mu ( \{ x: \mid f_n(x)+g_n(x)-f(x)-g(x) \mid \geq \epsilon \} ) \leq$ $\mu ( \{ x: \mid f_n(x)-f(x) \mid \geq \frac { \epsilon }{2} \} ) + \mu ( \{ x: \mid g_n(x)-g(x) \mid \geq \frac { \epsilon }{2} \} ) \rightarrow 0$

Therefore proved (a). Is this right?

For (b), I know that I need to show that:

$\mu ( \{ x: \mid f_ng_n(x)-fg(x) \mid \geq \epsilon \} ) \rightarrow 0$

But umm... a bit stuck here, how should I incorporate the fact that $\mu (X) < \infty$ ?

Thank you!

**tonio** · November 15th 2009, 12:20 AM

Originally Posted by tttcomrader

Suppose $f_n \rightarrow f$ in measure, and $g_n \rightarrow g$ in measure, prove that (a) $f_n+g_n \rightarrow f+g$ in measure and that (b) $f_ng_n \rightarrow fg$ in measure if $\mu (X) < \infty$ , not if $\mu (X) = \infty$

Proof so far.

(a) I think this is pretty easy.

I know that $\forall \epsilon > 0$ , we have $\mu ( \{ x : \mid f_n(x)-f(x) \mid \geq \epsilon ) \rightarrow 0$ and $\mu ( \{ x : \mid g_n(x)-g(x) \mid \geq \epsilon ) \rightarrow 0$ .

Now, since $\{ x: \mid f_n(x)+g_n(x)-f(x)-g(x) \mid \geq \epsilon \}$ $\subset \{ x: \mid f_n(x)-f(x) \mid \geq \frac { \epsilon }{2} \} \cup \{ x: \mid g_n(x)-g(x) \mid \geq \frac { \epsilon }{2} \}$

Question 1: Is this really right? I took this from a proof in the book, but I'm failing to fully understand why it is true.

So it follows that $\mu ( \{ x: \mid f_n(x)+g_n(x)-f(x)-g(x) \mid \geq \epsilon \} ) \leq$ $\mu ( \{ x: \mid f_n(x)-f(x) \mid \geq \frac { \epsilon }{2} \} ) + \mu ( \{ x: \mid g_n(x)-g(x) \mid \geq \frac { \epsilon }{2} \} ) \rightarrow 0$

Therefore proved (a). Is this right?

For (b), I know that I need to show that:

$\mu ( \{ x: \mid f_ng_n(x)-fg(x) \mid \geq \epsilon \} ) \rightarrow 0$

But umm... a bit stuck here, how should I incorporate the fact that $\mu (X) < \infty$ ?

Thank you!

$\left|f_n(x)g_n(x)-f(x)g(x)\right|=\left|f_n(x)g_n(x)-f_n(x)g(x)+f_n(x)g(x)-f(x)g(x)\right|\leq$ $\left|f_n(x)\right|\left|g_n(x)-g(x)\right|+\left|f_n(x)-f(x)\right|\left|g(x)\right|$

But $\mu(\{x\,:\,|f(x)|\geq \epsilon\})\leq \mu (X)<\infty\,\,\,\forall\,\,measurable\,\,function\ ,\,f(x)$ ...

Tonio

**ramdayal9** · April 25th 2010, 08:44 AM

hi, im trying to do this proof too.
Can you explain how to go from here?

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