# Chicken-Egg problem with Fubini’s theorem

Fubini's theorem states that if you have a double integral over a function $f(x,y)$, then you can compute the integral as an iterated integral, if $f(x,y)$ is in $\mathcal{L}^1$. But to find out if $f$ is in $\mathcal{L}^1$ you need to compute the double integral.

What am I missing? The examples I found all apply Fubini's theorem without checking that $f(x,y)$ is in $\mathcal{L}^1$. Many thanks for any clarification!

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If your function $f$ is measurable you can consider $|f|$ and apply Tonelli's theorem to show that $\int |f| d(\mu \times \nu) < \infty$. Then you can use Fubini's theorem to conclude that $\int \int f d \mu d \nu = \int \int f d \nu d \mu$. –  Nuno Jan 13 '11 at 16:38
You're right. It is a very rare situation that you know in advance that a function is integrable with respect to the product measure - it is often hard enough to show that a given function is measurable on the product spaces. My take on this is that Fubini's theorem is simply the best possible theoretical result on product measures and in the vast majority of the situations you only have to deal with $\sigma$-finite measure spaces anyway, so that Tonelli applies and the trick suggested by Nuno (or some variation of it) works. –  t.b. Jan 13 '11 at 17:54
You don't have to compute the double integral; you just have to bound it. For example if both of the underlying measure spaces are finite and f is bounded then this is automatically true. Similarly if you are computing an integral over R^2 it suffices to bound f on a sequence of compact subsets of R^2, say concentric circles or unit squares. –  Qiaochu Yuan Jan 13 '11 at 17:55
Thank you, Qiaochu Yuan. I would like to accept your comment as the answer to my question, if you don't mind re-posting it as answer. –  Matt N. Jan 13 '11 at 20:04

For Fubini to apply you need $f$ to be in $L^1$ as a function of $x$ for almost every $y$ and as a function of $y$ for almost every $x$.

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No, not really. This is a consequence of $f \in L^{1}(X \times Y)$ (at least if $X$ and $Y$ are complete). If $X$ and $Y$ fail to be complete then it is a bit more subtle, but in essence still true. See e.g. the books of Halmos or Royden –  t.b. Jan 13 '11 at 17:47
That's right. Thanks for the correction. –  Joe Johnson 126 Jan 13 '11 at 18:31