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ANALYSIS II: integration

  • Thread starter jjou
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Let [tex]A[/tex] be a rectangle in [tex]\mathbb{R}^k[/tex], [tex]B[/tex] be a rectangle in [tex]\mathbb{R}^n[/tex], [tex]Q=AxB[/tex]. If [tex]f:Q\rightarrow\mathbb{R}[/tex] is a bounded function and [tex]\int_Q f[/tex] exists, show that [tex]\int_B f(x,y)[/tex] exists for every [tex]x\in A-D[/tex] where [tex]D[/tex] is a set of measure zero in [tex]\mathbb{R}^n[/tex].



I know that, for fixed [tex]x\in A[/tex], [tex]\int_B f(x,y)[/tex] exists if [tex]D_x=\{y\in B|f\mbox{ is discontinuous at }(x,y)\}[/tex] has measure zero.

A set has measure zero if, for any [tex]\epsilon>0[/tex], there exists a covering of that set by countably many rectangles [tex]D_1, D_2, ...[/tex] such that [tex]\sum_i v(D_i)<\epsilon[/tex] (where [tex]v(D_i)[/tex] is the volume of [tex]D_i[/tex]).



My idea was to let [tex]D[/tex] be the set of [tex]x\in A[/tex] such that [tex]D_x[/tex] does not have measure zero. Then, clearly, we have that [tex]\int_B f(x,y)[/tex] exists for [tex]x\in A-D[/tex]. Then I would only have to show that [tex]D[/tex] has measure zero. I would do this by showing that [tex]f[/tex] is not continuous at points in [tex]D[/tex] (obvious). Since [tex]f[/tex] is integrable over [tex]Q[/tex], [tex]D[/tex] must have measure zero in [tex]Q[/tex].



My problem is that I know that a set [tex]D[/tex] which does not have measure zero in [tex]A[/tex] might have measure zero in [tex]Q[/tex] (since [tex]Q[/tex] is in a higher dimension). Is this not the right approach?
 

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