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Suppose we have two probability measures $\mu$ and $\delta$ on $(X, \mathcal{B})$ such that $ \delta <<\mu << \delta $. How can I prove that $f \in L^1(X,\mathcal{B}, \mu)$ iff $f \in L^1(X,\mathcal{B}, \delta)$?

My idea was to use that the Radon Nikodym theorem. So we know there exist $g$ $\mu$-measurable and $g^{-1}$ $\delta$-measurable such that for all $B \in \mathcal{B}$:

$$\delta(B) = \int_B g d \mu$$

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and

$$\mu(B) = \int_B g^{-1} d \delta$$.

Then if $f$ is in $L^1(X,\mathcal{B}, \mu)$, we have

$$\int_X |f| d\delta = \int_X |f g| d\mu$$

and so $f \in L^1(X,\mathcal{B}, \delta)$ if we can claim that $fg \in L^1(X,\mathcal{B}, \delta)$. However, we only know that $g$ is $\mu$-measurable! So I am not sure if this works… Any thoughts? Of course one can use a Holder type inequality for probability spaces to conclude that $\int_X |f g| d\mu \leq \int_X |f| d\mu \int_X |g| d\mu$ but can I say that $\int_X |g| d\mu < \infty$??

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Your claim is not true.

Take $X = (0,1)$, $\mu = m$ Lebesgue measure, and $\delta$ defined by $d\delta = g d\mu$ where $g(x) = \frac{1}{2} x^{-1/2}$. Then $\mu, \delta$ are both probability measures and are mutually absolutely continuous.

However, $g$ is in $L^1(\mu)$ but not in $L^1(\delta)$, since $\int g\,d\delta = \int g^2\,dm = \infty$.

It is true that $\int |g|\,d\mu = 1$, but the “Holder type inequality” you claim is false. In the above example, take $f=g$; then $\int |fg|\,d\mu = \infty$ while $\int |f|\,d\mu = \int |g| \,d\mu = 1$.

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