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Thread: Sum and product of functions convergence in measure

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    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
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  3. #3
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    hi, im trying to do this proof too.
    Can you explain how to go from here?
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