# show that there exists $f$ s.t. $\int_E(1-f)d\mu=\int_Efd\nu$

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 $$\mu\ll\psi$$?

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

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
• 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). Oct 30, 2019 at 16:38
• 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: $f \cdot (\mu + \nu)=f\cdot. [(\mu^+-\mu^-)+((\nu^+-\nu^-))] = f\cdot. [(\mu^++\nu^+)-f\cdot.((\mu^-+\nu^-))]$ Oct 30, 2019 at 16:40
• I do not understand what you are asking. Certainly my $f$ and my $h$ need not be equal. Oct 30, 2019 at 16:49
• 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. Oct 30, 2019 at 16:49