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I am confused with how to change the variables of the double integral below after invoking Fubini's Theorem; I am not very familiar with the wedge 'meet' notation which seems to arise. How is it that on changing the limits of the double integral

$$\int_{-\infty}^t\int_x^\infty f(y)g(x,y)\text{d}y\,\text{d}x$$

we obtain

$$\int_{-\infty}^{\infty}\int_{\infty}^{t\wedge y} f(y)g(x,y)\text{d}x\,\text{d}y?$$

I have had a look at the question Fubini's Theorem and Integral Bounds and tried to follow the reasoning they used for changing the limits there. With that in mind, in our initial region is

$$\{(x,y):x\in(-\infty,t)\}\,\bigcap\,\{(x,y):y\in(x,\infty)\}.$$

I can see that the new $y$-region component will be $\{(x,y):y\in(-\infty,\infty)\}$, since we seen that $x$ originally varies from $-\infty$. The issue for changing the $x$-region component is that it doesn't involve $y$ explicitly; how is it that we then have as the new $x$-region $\{(x,y):x\in(-\infty,t\wedge y)\}$? And what does the '$\wedge$' mean exactly in this context?

Jeremy Jeffrey James
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Just sketch the integration region and change the order of integration 'graphically'. Then you will notice that the upper bound for the $x$-variable is $\min (t,y)$. So $t \wedge y = \min(t,y)$.

daw
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    As a side note: the notation $a \wedge b = \min { a, b }$ isn't rare in measure theory, as well as $a \vee b = \max { a, b }$. – Adayah Apr 30 '19 at 10:26
  • I'll need to bear these in mind then; thanks! In the texts I have been going through so far I haven't encountered it until now. – Jeremy Jeffrey James Apr 30 '19 at 12:07