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Math Help - Sum and product of functions convergence in measure

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

    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!
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    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
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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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