Let $\mu$ and $\nu$ be finite measures on a measure space $(X,\mathcal{A})$. Show that there is a nonnegative measurable function $f$ on $X$ such that for all $E \in \mathcal{A}$, $$\int_E(1-f)d\mu=\int_Efd\nu$$

I have no clue how to show the above:

I started using below but I stuck

$$\mu(E)=\int_Efd\nu + \int_Efd\mu = \int_Efd(\nu+\mu)$$

defining $\psi=\nu+\mu$ , I have to show that $\mu\ll\psi$?

  • $\begingroup$ Yes, that will do. Then you can apply the Radon-Nikodym theorem. $\endgroup$
    – drhab
    Oct 30, 2019 at 16:26
  • $\begingroup$ yes, for positive measure, (as Will suggested below) was trivially correct. But I was thinking the measures are signed measure. $\endgroup$
    – domath
    Oct 30, 2019 at 16:35

1 Answer 1


You want to show $\mu = f \cdot (\mu + \nu)$ for some $f.$

If the result were true for positive $\mu$ and $\nu,$ then $\mu^+ = f \cdot (\mu^+ + \nu^+)$ and $\mu^- = g \cdot (\mu^- + \nu^-),$ then $\mu = \mu^+ - \mu^- = f \cdot \mu^+ - g \cdot \mu^- + f \cdot \nu^+ - g \cdot \nu^-.$ Since $\mu^+ \perp \mu^-,$ it turn outs $fg = 0$ (one of them is always zero). Thus, $h = f - g$ satisfies $f \cdot \mu^+ = h \cdot \mu^+$ and similarly for $\mu^-, \nu^\pm.$ Hence, $\mu = h \cdot (\mu + \nu).$ Thus, suffices to show the result for positive measures.

The result is trivial for positive measures for you can do what you did already and the relation $\mu(N) \leq \psi(N)$ shows that $\mu$ is absolutely continuous relative to $\psi.$ Q.E.D

  • $\begingroup$ I just realised that I did not showed that $h \geq 0.$ In fact, that restriction may be hard to prove for signed measures (I believe it should be false, take $\nu$ a negative measure and $\mu$ positive). $\endgroup$
    – William M.
    Oct 30, 2019 at 16:38
  • $\begingroup$ is this $f$ in $\mu^+ = f \cdot (\mu^+ + \nu^+)$ the same as $f$ in $\mu = f \cdot (\mu + \nu)$? I saw as you did this: $f \cdot (\mu + \nu)=f\cdot. [(\mu^+-\mu^-)+((\nu^+-\nu^-))] = f\cdot. [(\mu^++\nu^+)-f\cdot.((\mu^-+\nu^-))]$ $\endgroup$
    – domath
    Oct 30, 2019 at 16:40
  • $\begingroup$ I do not understand what you are asking. Certainly my $f$ and my $h$ need not be equal. $\endgroup$
    – William M.
    Oct 30, 2019 at 16:49
  • 1
    $\begingroup$ My $f$ and your $f$ are also not equal, but you need to understand $f$ us just a symbol that represents a function, I used $f,$ $g$ and $h$ in place of $f$ depending on what measure was under consideration. $\endgroup$
    – William M.
    Oct 30, 2019 at 16:49

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