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Convergence in measure

  1. Jan 4, 2012 #1

    Fredrik

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    Not sure where to post about measure theory. None of the forums seems quite right.

    Suppose that ##(X,\Sigma,\mu)## is a measure space. A sequence ##\langle f_n\rangle_{n=1}^\infty## of almost everywhere real-valued measurable functions on X is said to converge in measure to a measurable function f, if for all ε>0, ##\mu(\{x\in X:|f_n(x)-f(x)|\geq\varepsilon\})\to 0##.

    The book says that it's easily seen that if ##\langle f_n\rangle## converges in measure to two measurable functions f and g, then f=g almost everywhere. I don't see it.

    I understand that I need to prove that ##\mu(\{x\in X: |f(x)-g(x)|>0\})=0##, but I don't even see how to begin.
     
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  3. Jan 4, 2012 #2

    micromass

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    Note that

    [tex]|f(x)-g(x)|\leq |f(x)-f_n(x)|+|f_n(x)-g(x)|[/tex]

    So if [itex]|f(x)-g(x)|\geq \varepsilon[/itex], then also

    [tex]|f(x)-f_n(x)|+|f_n(x)-g(x)|\geq \varepsilon[/tex].

    But this must mean that either [itex]|f(x)-f_n(x)|\geq \varepsilon/2[/itex] or [itex]|f_n(x)-g(x)|\geq \varepsilon/2[/itex] (proceed by contradiction: assume that neither holds). So

    [tex]\{x\in X~\vert~|f(x)-g(x)|\geq \varepsilon\}\subseteq \{x\in X~\vert~|f(x)-f_n(x)|\geq \varepsilon/2\}\cup \{x\in X~\vert~|g(x)-f_n(x)|\geq \varepsilon/2\}[/tex]

    Taking [itex]\mu[/itex] of both sides yields

    [tex]\mu\{x\in X~\vert~|f(x)-g(x)|\geq \varepsilon\}\leq \mu\{x\in X~\vert~|f(x)-f_n(x)|\geq \varepsilon/2\}+ \mu\{x\in X~\vert~|g(x)-f_n(x)|\geq \varepsilon/2\}[/tex]

    I think you can take it from here. It's just a limiting argument now.

    Another nice argument (but a little more difficult to prove) is to show that the sequence has a subsequence which converges a.e. Now f=g a.e. by uniqueness of the limit a.e.
     
  4. Jan 4, 2012 #3

    I like Serena

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    From wikipedia:

    "Calculus (Latin, calculus, a small stone used for counting) is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This subject constitutes a major part of modern mathematics education."

    and:

    "Mathematical analysis, which mathematicians refer to simply as analysis, has its beginnings in the rigorous formulation of infinitesimal calculus. It is a branch of pure mathematics that includes the theories of differentiation, integration and measure, limits, infinite series,[1] and analytic functions."


    Seems to me this fits Calculus & Analysis.
     
  5. Jan 4, 2012 #4

    micromass

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    I prefer this in the probability forum. Measure theory (and convergence in probability, like here) are really important probability concepts. So it might make sense that a probabilist knows more about them then an analyst (it certainly is my experience that this is the case).
     
  6. Jan 4, 2012 #5

    Fredrik

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    Ah, I think I understand it now. Your argument tells us that for all ##\varepsilon,\varepsilon'>0##, we have ##\mu(\{x\in X:|f(x)-g(x)|\geq\varepsilon)<\varepsilon'##. This implies that for all ##\varepsilon>0##, we have ##\mu(\{x\in X:|f(x)-g(x)|\geq\varepsilon\})=0##. We still have to do something tricky for the final step, something like this:

    $$\big\{x\in X:|f(x)-g(x)|>0\big\}=\bigcup_{n=1}^\infty\bigg\{x\in X:|f(x)-g(x)|\geq\frac{1}{n}\bigg\}$$
    $$\mu\big(\big\{x\in X:|f(x)-g(x)|>0\big\}\big) \leq\sum_{n=1}^\infty\mu\bigg( \bigg\{x\in X:|f(x)-g(x)|\geq\frac{1}{n}\bigg\} \bigg)=0$$ Thank you.
     
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