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I am going through exercise problems of Rudin, and I came across this question.

Find an example of a positive continuous function $f$ in the open unit square in $\mathbb{R}^2$, whose integral (relative to Lebesgue measure) is finite but such that $\int_0^1 f_x(y) dy$ is infinite for some $x \in (0,1)$. (Here, $f_x(y) = f(x,y)$.)

I tried thinking of cases when Fubini theorem fails, for example $\displaystyle f(x,y) = \frac{2x-1}{y}$, but I found it difficult to evaluate the integral in Lebesgue sense.

Any help is greatly appreciated.

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up vote 1 down vote accepted

Edit: Corrected typo in the definition of $f$.

Consider the function $$ f(x,y)= f_x(y) = \frac{1}{y^{1-\sqrt{|x-1/2|}}} $$ We then have $$ \int_0^1 f_x(y)dy = \begin{cases} \frac{1}{\sqrt{|x-1/2|}} & x\neq 1/2 \\ \infty & x=1/2 \end{cases} $$ and

$$ \int_0^1 \left[\begin{cases} \frac{1}{\sqrt{|x-1/2|}} & x\neq 1/2 \\ \infty & x=1/2 \end{cases}\right]dx = 2\sqrt{2}. $$

Tonelli's theorem implies that the Lebesgue integral $\int_{[0,1]^2}f(x,y)dxdy$ is finite and equal to $2\sqrt{2}$, yet $\int_0^1 f_{1/2}(y)dy$ is infinite.

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I believe the above finite value was found by integrating with respect to $y$ first. Does it give the same value when integrated with respect to $x$ first? Or does that not matter? – aviness Dec 19 '12 at 11:16
If the integral is finite for one particular order of integration, it is finite for the other order as well. The value of the integral is the same in both cases and coincides with the two-dimensional integral (i.e. with respect to two-dimensional Lebesgue measure). In my answer I integrated with respect to $y$ first, and then with respect to $x$. – Eckhard Dec 19 '12 at 14:29

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