cindy asked in 科學數學 · 1 decade ago

實變問題(Fubini's Theorem)

Let f(x,y) be nonnegative and measurable in R^2.

Suppose that for almost every x belongs to R,

f(x,y) is finite for almost every y.

Show that for almost every y belongs to R, f(x,y) is finite for almost every x.

1 Answer

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  • 1 decade ago
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    Claim:Let E be measurable set of |R^2, Suppose for almost every x€|R

    {y:(x,y)€E} has |R measure zero,then for almost every y€|R

    {x:(x,y)€E} has measure zero

    proof of claim:

    Take χ_E,the characteristic function,By Tonell's Theorem

    ∫∫χ_Edxdy=∫∫χ_Edydx

    ∫∫χ_Edxdy=0(因為∫χ_Edy=0)

    =>∫χ_Edx=0 for almost every y

    =>|{x:(x,y)€E}|=0 for almost every y

    Hence for almost every y€|R,{x:(x,y)€E} has measure zero

    Now let E={(x,y)€|R^2:f(x,y)=+∞},then for almost every x€R

    {y:(x,y)€E} has measure zero

    So by our claim,{x:(x,y)€E} has measure zero for almost every y

    That is for almost every y, f(x,y) is finite for almost every x€|R

    2008-06-16 23:06:34 補充:

    這題是用到Tonells Theorem,不是Fubinis Theorem

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