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Prove f>=0 almost everywhere

  1. Dec 14, 2011 #1
    1. The problem statement, all variables and given/known data
    If {f_n} is a collection of measurable functions defined on R and satisfying: |f_n(x)|<= 1 for all n in N and x in R and f_n(x) >=0 almost everywhere on R for all n in N and f(x) = inf{f_n(x)|n in N}, then f(x) >= 0 almost everywhere on R.


    2. Relevant equations

    Almost everywhere, sequences of functions, measurability


    3. The attempt at a solution

    Assume {f_n} is a collection of measurable functions defined on R satisfying the criteria above. Since for all n in N f_n(x) >= 0 a.e on R, define S_n = {x | f_n(x) < 0}. Then by definition for all n in N m(S_n) = 0. Now let [tex] S = \bigcup_{n=1}^\infty S_n. [/tex] Since S is the union of several sets S_n with measure of zero, then m(S) = 0. So, we know f(x) = inf{f_n(x) | n in N} >= 0 for all x in R-S. But m(S) = 0, so f(x) >= 0 almost everywhere, as desired.

    I'm wondering if I can necessarily state that the inf of the f_n's will be >= 0. I'm not that great at catching little details like that, so having someone give the proof a quick look-over would be incredibly helpful.

    Thanks!
     
  2. jcsd
  3. Dec 14, 2011 #2

    Office_Shredder

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    It helps to clarify things if you fix x to be a specific value, say b. If [itex]f_n(b) \geq [/itex] for all values of n, then is [itex] \inf\{f_n(b)} \geq 0[/itex]? This follows immediately from the definition of infimum.

    Then conclude that S can only consist of values of x for which at least one fn(x)<0
     
  4. Dec 15, 2011 #3
    Ah! Thanks for the input!
     
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