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 $$ \pi(\mathsf{A}) = c\mu(\mathsf{A}) \qquad \forall\,\mathsf{A}\in\mathcal{X} $$

  • 1
    $\begingroup$ If both $\pi$ and $\mu$ are probability measures, then $c$ must be $1$. In other words, $\pi \equiv \mu$. $\endgroup$
    – Zhanxiong
    Commented Jul 12, 2023 at 1:30

1 Answer 1


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.)


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