## Trending News

# 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

- Scharze spaceLv 71 decade agoFavorite Answer
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