# Does a constant Radon-Nikodym derivative imply the measures are multiples of each other?

Let $$(\mathsf{X}, \mathcal{X})$$ be a measurable space and $$\pi$$ and $$\mu$$ be two probability measures on it such that $$\pi \ll \mu$$ with constant Radon-Nikodym derivative $$\frac{d\pi}{d\mu} = \text{constant}$$

Does this imply that the measures are multiples of each other? I.e. that there exists a constant $$c > 0$$ such that $$\pi(\mathsf{A}) = c\mu(\mathsf{A}) \qquad \forall\,\mathsf{A}\in\mathcal{X}$$

• If both $\pi$ and $\mu$ are probability measures, then $c$ must be $1$. In other words, $\pi \equiv \mu$. Jul 12 at 1:30

## 1 Answer

Yes, that is correct. From the definition of what the Radon-Nikodym derivative is (which is essentially a type of anti-integral), for any event $$\mathcal{A} \in \mathscr{X}$$ and any measures $$\pi$$ and $$\mu$$ obeying the requirements you have given, we must have:

\begin{align} \pi(\mathcal{A}) &= \int \limits_\mathcal{A} \frac{d\pi}{d\mu} \ d \mu \\[6pt] &= \int \limits_\mathcal{A} \text{constant} \ d \mu \\[6pt] &= \text{constant} \times \int \limits_\mathcal{A} \ d \mu \\[12pt] &= \text{constant} \times \mu(\mathcal{A}), \\[6pt] \end{align}

which shows that the measures are proportionate in this case, with the constant of proportionality equal to the constant value for the Radon-Nikodym derivative. If $$\pi$$ and $$\mu$$ are both probability measures then they must both obey the norming axiom of probability, which means they must both have a total measure of one. In this case you have $$\text{constant} = 1$$ and so $$\pi = \mu$$ (i.e., they are the same probability measure). (This is why when we work with probability, showing proportionality is sufficient to show that two distributions are the same.)