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Hi everyone: Suppose $B_{1}$ and $B_{2}$ are balls in $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$ (let's say for $m,n\geq2)$. Suppose that $f(x,y)$ is defined and measurable eveywhere. Beside, $$0\leq \int_{B_{1}}f(x,y)dx\leq C$$ for all $y$ and $C$ is a constant. Does this imply that $f$ is integable, or integrable in the large sense (i.e., at least one of the positive or negative parts is integrable)? Thanks for your help.

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  • $\begingroup$ You mean if $\int_{B_2} \int_{B_1} f(x,y) dx dy < \infty $ ? $\endgroup$ – Adam May 29 '16 at 1:49
  • $\begingroup$ No. I mean $\int_{B_{2}}\int_{B_{1}}\mid f(x,y)\mid d\lambda(x)d\lambda(y)<+\infty$. Or at least the double integral of $f^{+}$ (or $f^{-}$) is finite. $\endgroup$ – M. Rahmat May 29 '16 at 4:22
  • $\begingroup$ Yeah sorry I forgot, ofc. a function is called Lebesgue integrable when $\int |f| dx < \infty $ and not $\int f dx < \infty$. $\endgroup$ – Adam May 29 '16 at 10:00
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This does not imply that $f$ is integrable.

Counter example

Consider $f(x,y) = \begin{cases} \frac{xy(y^2 -x^2)}{(x^2 + y^2)} & \text{for } y \in (0,2] \\ 0 & else \end{cases}$ and $B_1= [0,1]$ and $B_2=[0,2]$.

Then

$$0 \leq \int_0^1 f(x,y) \, dx = \frac{y}{2( y^2 +1)^2} \leq 1 $$ for all $y \geq 0$ and for $y < 0$ the integral is 0. In short, $f$ fulfills your condition.

However, computation reveals (see here) that

$$ \int_0^2 \int_0^1 f(x,y) \, dx \, dy = \frac{1}{5} $$ and

$$ \int_0^1 \int_0^2 f(x,y) \, dy \, dx = - \frac{1}{20} $$

and since

$$ \int_0^2 \int_0^1 f(x,y) \, dx \, dy \neq \int_0^1 \int_0^2 f(x,y) \, dy \, dx .$$

Fubinis Theorem implies that $f$ is not integrable.

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