# Thread: Sum and product of functions convergence in measure

1. ## Sum and product of functions convergence in measure

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

Proof so far.

(a) I think this is pretty easy.

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

Now, since $\displaystyle \{ x: \mid f_n(x)+g_n(x)-f(x)-g(x) \mid \geq \epsilon \} $$\displaystyle \subset \{ x: \mid f_n(x)-f(x) \mid \geq \frac { \epsilon }{2} \} \cup \{ x: \mid g_n(x)-g(x) \mid \geq \frac { \epsilon }{2} \} Question 1: Is this really right? I took this from a proof in the book, but I'm failing to fully understand why it is true. So it follows that \displaystyle \mu ( \{ x: \mid f_n(x)+g_n(x)-f(x)-g(x) \mid \geq \epsilon \} ) \leq$$\displaystyle \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:

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

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

Thank you!

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

Proof so far.

(a) I think this is pretty easy.

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

Now, since $\displaystyle \{ x: \mid f_n(x)+g_n(x)-f(x)-g(x) \mid \geq \epsilon \} $$\displaystyle \subset \{ x: \mid f_n(x)-f(x) \mid \geq \frac { \epsilon }{2} \} \cup \{ x: \mid g_n(x)-g(x) \mid \geq \frac { \epsilon }{2} \} Question 1: Is this really right? I took this from a proof in the book, but I'm failing to fully understand why it is true. So it follows that \displaystyle \mu ( \{ x: \mid f_n(x)+g_n(x)-f(x)-g(x) \mid \geq \epsilon \} ) \leq$$\displaystyle \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:

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

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

Thank you!

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

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

Tonio

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