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Measure theory, negation of equal almost everywhere

  1. May 1, 2014 #1
    If f=g a.e
    f and g are equal except at a measurable set with measure zero



    If two functions are not equal a.e what will then the negation be? Will there have to exist a set that is measurable, and f is not equal to g on this set, and this set has not measure 0?

    Or will the entire set where they are not equal be measurable, and have measure different from 0?
     
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  3. May 1, 2014 #2

    LCKurtz

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    There will exist a set of positive measure on which ##f\ne g##.
     
  4. May 1, 2014 #3
    Do the functions have to be different for all the values in that set, or is it enough that they are just different in one?
     
  5. May 1, 2014 #4

    Dick

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    Since your orignal statement can be phrased as "the set where ##f \ne g## is a measureable set of measure zero", the logical negation is simply that "the set where ##f \ne g## is NOT a measureable set of measure zero". To say more about that set you would need to know something about f and g and the nature of the measure. To say "the set where ##f \ne g##" sort of implies f and g differ on all points in the set.

    Since the real question seems to be, "Will there have to exist a set that is measurable, and f is not equal to g on this set, and this set has not measure 0?", the answer is no. Look at Vitali sets. That lets you split the unit interval up into an infinite number of nonmeasurable disjoint sets. If each set contained a measurable subset of positive measure, the measure of the unit interval would be infinite. That's not true in the usual notion of Lesbegue measure on the real line.
     
    Last edited: May 1, 2014
  6. May 2, 2014 #5
    Maybe I should have written that f and g also should be measurable. Then the set {x: f(x)[itex]\ne[/itex]g(x)} is always measurable. And if they are equal a.e, this set has measure 0, if they are not equal a.e. this set has measure larger than 0?

    And the same is true for a function that is finite a.e or 0 a.e? The sets {x: f(x) ≠ 0} and {x:f(x)=+-∞} are always measurable. So the negation only becomes that these sets have a measure larger than 0?
     
  7. May 2, 2014 #6

    Dick

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    Right. If a set is measureable and its measure isn't zero, then it has nonzero measure. Clearly. Still that's not really the logical negation of your statement - it requires some extra work (like proving the set is measureable) to conclude that.
     
  8. May 2, 2014 #7
    Yeah, I will try to be precise. And ask questions so that I can understand this fully.

    Let f and g be measurable functions and equal almost everywhere. Then
    {x: f(x)≠g(x)} is measurable, AND has measure zero.

    The logical negation of this is:
    {x: f(x)≠g(x)} is not measurable OR {x: f(x)≠g(x)} is measurable and has measure NOT zero.

    Since f and g is measurable we have that f-g is measurable and the set {x: f-g≠0}={x:f(x)≠g(x)}is measurable. So the negation now only is that
    {x: f(x)≠g(x)} is measurable and has a measure larger than zero.

    So we do NOT have to say that there EXIST a measurable set, where the condition fails has measure not zero. We know that THE entire set where it fails is measurable, and has measure zero?
     
  9. May 2, 2014 #8

    Dick

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    Yes, given those assumptions, if f and g are measureable and they are not equal a.e. then the set where ##f \ne g## is measurable and has nonzero measure.
     
  10. May 2, 2014 #9
    Thanks, and I thought negation was easy!.... :)
     
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