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Let $\mu$ and $\nu$ be two measures on a space $X$. Suppose

$$ \int_X fd\mu \int_X gd\nu = \int_X fd\nu \int_X gd\mu$$

for any integrable functions $f,g$. I would like to show that this implies that $\mu = \lambda\nu$ for some $\lambda > 0$, but I don't know how. Any help would be appreciated!


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up vote 9 down vote accepted

By setting $f=1_A, g=1_B$ you have $\mu A \, \nu B = \mu B \, \nu A$.

So, if there exists a set $B$ that satisfies $0 < \nu(B) < \infty$, then you have $\mu A =\frac{\mu B}{\nu B} \nu A$.

Addendum: The result is true generally, not that the additional generalization is particularly useful. Here is the proof anyway.

If there exists a set $B$ such that either $0 < \mu B < \infty$ or $0 < \nu B < \infty$, then the above shows the existence of a suitable constant.

If $\mu B = 0$ for all $B$, then the constant $\lambda =0$ will do, of course. Similarly for $\nu$.

So, suppose that for all suitable $B$, we have $\mu B = 0 $ or $\mu B = \infty$, similarly $\nu B = 0 $ or $\nu B = \infty$, and neither $\mu$ nor $\nu$ are identically $0$.

Now suppose $\mu A = 0$. Then $\mu B \, \nu A = 0$ for all suitable $B$. Since there is at least one $B$ such that $\nu B \neq 0$, we have $\nu A = 0$. Similarly for $\nu$. Hence $\mu A = 0 $ iff $\nu A = 0$. Since the only other possible value is $\infty$, we have that for all suitable sets $A$, $\mu A = \nu A$, and the constant $\lambda=1$ will do.

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Should be $\nu A$ at the end. – Ayman Hourieh Dec 13 '12 at 21:11
@AymanHourieh: Thanks! – copper.hat Dec 13 '12 at 21:11

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