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I have proven a problem but I am unsure whether it is correct because the proof seems so simple that I think I might be mistaken. Please be kind to comment on my proof and tell me whats wrong with it.

The problem: Let $A$ be a rectangle in $\Bbb R^k$ and B a rectangle in $\Bbb R^n$. Let $Q=A\times B$. Let $f:Q\to R$ be bounded and Riemann integrable. Show that if $\int_Qf$ exists, then $\int_yf(x,y)$ exists for $x\in A-D$ where D is a set of measure zero in $\Bbb R^k$.

My idea: It just seems like a direct application of Fubini theorem. I wrote that if $\int_Qf $ exists, then by fubini theorem, $\int_Qf=\int_x\int_y f$ and it implies that $\int_yf(x,y)$ exists.

Fubini's theorem version that I use: Let $A$ be a rectangle in $\Bbb R^k$ and B a rectangle in $\Bbb R^n$. Let $Q=A\times B$. Let $f:Q\to R$ be bounded. $f$ is in the form $f(x,y)$ for $x\in A, y\in B$. If f is integrable over Q, then $$\int_Q f=\int_x \int_yf=\int_y\int_xf$$

Is this right?

Thank you very much for your help

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  • Just to be sure, are you considering Riemann integrals? – Luiz Cordeiro Nov 27 '14 at 00:57
  • @LuizCordeiro Thank you. Yes. Is my approach right then? – Marion Crane Nov 27 '14 at 01:15
  • Apparently your approach is correct. Note that in your version of Fubini's Theorem, you only assume that $f$ is integrable over $Q$ and then conclude that $\int_Q f=\int_B\left[\int_Af(x,y)dx\right]dy$. For this to make sense, you have to assume that $\int_B f(x,y)dy$ is defined at least for almost every $x\in A$, which is what you want. (The "correct" way to enunciate Fubini's Theorem would be a bit more complicated, see http://math.stackexchange.com/a/310841/58818) – Luiz Cordeiro Nov 28 '14 at 17:31

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