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Math Help - measurable functions

  1. #1
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    measurable functions

    Let f and g be measurable functions from X to the extended reals. Prove that the sets

    A={x:f(x)<g(x)}
    and
    B={x:f(x)=g(x)}
    are measurable.

    And then prove that the set of points at which a sequence of measurable real valued functions converges is measurable.

    To show that A is measurable I am trying to work with intersections. I know that f is measurable iff {(a,inf]} is measurable for all a in R. Similarly for g. Can I simply take the intersection of such sets to get A? Is the intersection of measurable sets measurable?

    Can I do something similar for B?

    Assuming A and B are measurable, I suppose I have to use that to show the second part of the question, but I dont see how.
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  2. #2
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    1) Given the sigma algebra (the collection of measurable sets), A, then if X,Y\in A you have that X^c,Y^c\in A. Next X\cap Y=\left(X^c\cup Y^c\right)^c. So X\cap Y\in A.

    2) Let E_(n,m,k)=\left\{x:|f_n(x)-f_m(x)|<\frac{1}{k}\right\}. Then \bigcap_{k=1}^\infty\bigcup_{N=1}^\infty\bigcap_{n  =N}^\infty\bigcap_{m=N}^\infty E(n,m,k) is measurable. Now just show it is the same as the set of points where f_n converges. Oh and you need to justify that E(n,m,k), and the final set (with the intersections and unions) are both measurable.
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  3. #3
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    For set A,try this A= \bigcup_{r\in Q}(\{x:f(x)<r\}\bigcap \{x:g(x)>r\}),Q is the set of rational numbers.
    set B can be expressed as \{x:f(x)\le g(x)\} \bigcap \{x:g(x)\le f(x)\}
    And then prove that the set of points at which a sequence of measurable real valued functions converges is measurable.
    The required set is \{x:lim inf  f_n(x)=lim sup  f_n(x)\} where it follows that lim inf   f_n and lim sup  f_n are measurable functions.
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