# An identity involving Radon-Nikodym derivatives

The following result was stated without proof in [HAL] (note 4 to section 32 "Derivatives of Signed Measures", p. 136).

If $\mu_0$, $\mu_1$, and $\mu_2$ are finite measures, and if

$$\begin{array}{lcl} f_1 & = & \frac{d\mu_0}{d\left(\mu_0+\mu_1\right)} \\ f_2 & = & \frac{d\mu_0}{d\left(\mu_0+\mu_2\right)} \\ f & = & \frac{d\mu_0}{d\left(\mu_0+\mu_1+\mu_2\right)} \end{array}$$

then we have $\left(\mu_0+\mu_1+\mu_2\right)$-a.e.,

$$f(x) = \begin{cases} \frac{f_1(x)f_2(x)}{f_1(x)+f_2(x)-f_1(x)f_2(x)} &\mbox{if }f_1(x)f_2(x)\neq0,\\ 0 &\mbox{if }f_1(x)=f_2(x)=0 \end{cases} \space\space\mbox{(Alexander)}$$

An intuitive, symbolic proof was suggested by postmortes in another thread:

Write

$$\begin{array}{lcl} d\mu_0 & = & f_1 d\left(\mu_0+\mu_1\right)\space\space\mbox{(Gary)} \\ d\mu_0 & = & f_2 d\left(\mu_0+\mu_2\right)\space\space\mbox{(Terrence)}\\ d\mu_0 & = & f d\left(\mu_0+\mu_1+\mu_2\right)\space\space\mbox{(Leon)} \end{array}$$

Multiply (Gary) and (Terrence) throughout by $f_2$ and $f_1$, respectively, to obtain

$$\begin{array}{lcl} f_2d\mu_0 & = & f_1f_2 d\left(\mu_0+\mu_1\right) \\ f_1d\mu_0 & = & f_1f_2 d\left(\mu_0+\mu_2\right) \end{array}$$

Add up to obtain

$$(f_1+f_2)d\mu_0= 2f_1f_2d\mu_0 + f_1f_2d\mu_2 + f_1f_2d\mu_1$$

Subtract $f_1f_2d\mu_0$ from both sides to obtain

$$\left(f_1 + f_2 - f_1f_2\right)d\mu_0 = f_1f_2d\left(\mu_0 + \mu_1 +\mu_2\right)$$

Finally divide both sides by $d\left(\mu_0+\mu_1+\mu_2\right)$ and use (Leon) to obtain (Alexander).

Unfortunately, i fail to see how to turn these symbolic manipulations into a rigorous argument. Any help will be welcome.

### References

[HAL] Halmos, Paul Richard. "Measure Theory". Springer-Verlag, 1974.

• (Alexander) is a bit strange: the two cases do not seem to divide the space into two pieces. What if only one of $f_1$ and $f_2$ is zero? // I think Exercise (2b) on page 130f should be useful: if $\int g \, d\nu = \int fg\,d\mu$ then $\nu(E) = \int_E \frac{f}{1-f}\,d\mu$. Commented Apr 22, 2013 at 14:01
• @Martin: Thanks. (1) It's not so strange, since $f_1, f_2$ can be assumed to take values in the interval $[0,1]$ and in this interval the equation $x_1+x_2-x_1x_2=0$ admits a single solution: $x_1=x_2=0$. (2) In what way should exercise (2b) on page 130 prove useful? Commented Apr 22, 2013 at 15:30

Everything is already more or less rigorous. The first step is just a reformulation of the definitions of the $f_i$. For example: $f_1$ is the density of $\mu_0$ with respect to $\mu_0 + \mu_1$, so for all measurable $A$ its measure $\mu_0(A)$ can be obtained by integrating $f_1$ with respect to $\mu_0 + \mu_1$ over $A$. »$f_1 \mathrm{d}(\mu_0 + \mu_1)$« is just a shorthand notation for the definition of a measure by the last sentence.

Then the next step follows from a theorem, which you can re-prove easily: if $\mu$ has density $f$ with respect to $\nu$ and $\lambda$ has density $g$ with respect to $\mu$, then $\lambda$ (which is $g \mathrm{d}\mu$ by what I've already said) has density $fg$ with respect to $\nu$ (i. e. $\mathrm{d}\lambda = fg \mathrm{d}\nu$).

The third step uses another theorem, which states that the sum of measures with density with respect to a common control measure has the sum of those densities as density with respect to the common measure. The right-hand sides just uses the definition of the sum of two measures.

The fourth step follows analogously, only that you should be careful not to subtract $\infty$ from $\infty$.

In the last step, you can substitute (Leon) first and get an equation of two measures with density with respect to $\nu := \mu_0 + \mu_1 + \mu_2$. At $\nu$-almost all points theses densities must coincide (otherwise integrate both densities over the set of points where equality does not hold to get a contradiction). Then it is just a matter of shuffling the terms around. Edit: For Martin's caveat: the comparison also gives that $f_1$ vanishes iff $f_2$ vanishes ($\nu$-almost everywhere).

• Thank you, Thomas. Regarding the third step. Did you invoke the result (mentioned in [HAL] p. 133) that if $\nu_1$, $\nu_2$ and $\mu$ are finite and $\nu_1,\nu_2\ll\mu$, then $\frac{d\left(\nu_1+\nu_2\right)}{d\mu}=\frac{d\nu_1}{d\mu}+\frac{d\nu_2}{d\mu}$? Cause if you did, i don't see how it applies in the present case. Can you perhaps rewrite the equations to make this deduction more readily visible? Commented Apr 22, 2013 at 15:26
• I've got it now. Thanks! Commented Apr 22, 2013 at 16:16
• Yes, that should be it. But you can also just use how the sum of two measures is defined: Set $\mathrm{d}\nu_1 = f_1 \mathrm{d}\mu_0$ and $\mathrm{d}\nu_2 = f_2 \mathrm{d}\mu_0$, then $\nu_1 + \nu_2 = (f_1 + f_2) \mathrm{d}\mu_0$. Commented Apr 22, 2013 at 16:31

The following answer is divided into three sections:

Auxiliary Results and Definitions
A collection of fourteen definitions and theorems that will be used in the proof. Only a selection of the theorems are proved. The rest are standard or easy to prove.

A Restatement of the Result to be Proved
A reformulation of Halmos' theorem in more precise terms, using the notation established in the first section.

Proof
A rigorous proof of Halmos' theorem. This section is further divided into two sub-sections: Existence and Non-Uniqueness. The scope of each sub-section is outlined in the beginning of the Proof section.

## Auxiliary Results and Definitions

All the results in this section, except propositions 11, 13 and 14, are either standard or easily proved. I will therefore prove only these three propositions.

1. Definition
1. Denote by $\overline{\mathbb{R}}$ the extended real line $\mathbb{R}\cup\left\{\pm\infty\right\}$.
2. Denote by $\mathfrak{B}$ the Borel field on the extended real line.
2. Definition Let $\Omega$ be a non-empty set and let $F,G\subseteq\Omega\rightarrow\overline{\mathbb{R}}$.
1. The product of $F$ and $G$ is the set $$FG:=\left\{fg:\mid\left(f,g\right)\in F\times G\right\}$$
2. The set of $\left(F,G\right)$-summable functions is $$\mathfrak{S}\left(F,G\right) := \left\{\left(f,g\right)\in F\times G\mid:\forall\omega\in\Omega,\space \left(f\left(\omega\right),g\left(\omega\right)\right)\neq\pm\left(\infty,-\infty\right)\right\}$$
3. The sum of $F$ and $G$ is the set $$F+G := \left\{f+g:\mid \left(f,g\right)\in\mathfrak{S}\left(F,G\right)\right\}$$
3. Definition Let $\left(\Omega_i,\mathcal{A}_i\right)$, $i=1,2$ be two measurable spaces. Then we denote the set of all $\mathcal{A}_1/\mathcal{A}_2$-measurable function by $\left(\Omega_1,\mathcal{A}_1\right)\rightarrow\left(\Omega_2,\mathcal{A}_2\right)$.
4. Definition Let $\left(\Omega,\mathcal{A}, \mu\right)$ be a measure space and let $f\in\left(\Omega,\mathcal{A}\right)\rightarrow\left(\overline{\mathbb{R}},\mathfrak{B}\right)$. We write $$f=g\space\space\left[\mathcal{A},\mu\right]$$ to indicate that $\mu\left(\left\{\omega\in\Omega\mid:f\left(\omega\right)\neq g\left(\omega\right)\right\}\right)=0$.
5. Proposition Let $\left(\Omega,\mathcal{A}, \mu\right)$ be a measure space. The relation $f=g\space\space\left[\mathcal{A},\mu\right]$ is an equivalence relation on $\left(\Omega,\mathcal{A}\right)\rightarrow\left(\overline{\mathbb{R}},\mathfrak{B}\right)$.
6. Definition Let $\left(\Omega,\mathcal{A}, \mu\right)$ be a measure space and let $f\in\left(\Omega,\mathcal{A}\right)\rightarrow\left(\overline{\mathbb{R}},\mathfrak{B}\right)$. We denote by $\left[f\right]_{\mathcal{A},\mu}$ the set $$\left\{g\in\left(\Omega,\mathcal{A}\right)\rightarrow\left(\overline{\mathbb{R}},\mathfrak{B}\right)\mid:g=f\space\space\left[\mathcal{A},\mu\right]\right\}$$
7. Proposition Let $\left(\Omega,\mathcal{A},\mu\right)$ be a measure space and let $f:\left(\Omega,\mathcal{A}\right)\rightarrow\left(\overline{\mathbb{R}},\mathfrak{B}\right)$. Suppose $0\leq f\left(\omega\right)$ for all $\omega\in\Omega$ or $f\left(\omega\right)\leq0$ for all $\omega\in\Omega$. Then $$\int_\Omega f d\mu=0\implies f=0\mathbb{1}_\Omega\space\left[\mathcal{A},\mu\right]$$
8. Theorem (Radon-Nikodym I) Let $S=\left(\Omega,\mathcal{A}\right)$ be a measurable space and let $\mu$, $\nu$ be $\sigma$-finite measures on $S$ such that $\nu\ll\mu$. Then there's a non-negative $f\in\left(\Omega,\mathcal{A}\right)\rightarrow\left(\overline{\mathbb{R}},\mathfrak{B}\right)$ such that for all $B\in\mathcal{A}$, $$\nu(B)=\int_B fd\mu$$
9. Definition (Radon-Nikodym Derivative) Let $S=\left(\Omega,\mathcal{A}\right)$ be a measurable space and let $\mu$, $\nu$ be $\sigma$-finite measures on $S$ such that $\nu\ll\mu$. We will denote by $\frac{d\nu}{d\mu}$ a certain fixed, but otherwise arbitrary, non-negative $f\in\left(\Omega,\mathcal{A}\right)\rightarrow\left(\overline{\mathbb{R}},\mathfrak{B}\right)$ that satisfies for all $B\in\mathcal{A}$, $$\nu(B)=\int_B fd\mu$$
10. Theorem (Radon-Nikodym II) Let $S=\left(\Omega,\mathcal{A}\right)$ be a measurable space and let $\mu$, $\nu$ be $\sigma$-finite measures on $S$ such that $\nu\ll\mu$. Then the set consisting of those $g\in\left(\Omega,\mathcal{A}\right)\rightarrow\left(\overline{\mathbb{R}},\mathfrak{B}\right)$ ($g$ need not be non-negative) that have a (not necessarily finite) integral w.r.t. $\left(\Omega,\mathcal{A},\mu\right)$ and such that for all $B\in\mathcal{A}$, $$\nu(B)=\int_B gd\mu$$ is equal to the set $\left[\frac{d\nu}{d\mu}\right]_{\mathcal{A},\mu}$.
11. Proposition Let $S=\left(\Omega,\mathcal{A}\right)$ be a measurable space and let $\nu$ and $\mu$ be $\sigma$-finite measures on $S$.

1. If for all $B\in\mathcal{A}$, $\nu(B)\leq\mu(B)$, then $\nu\ll\mu$ and there's some $f\in\left[\frac{d\nu}{d\mu}\right]_{\mathcal{A},\mu}$ such that $f\left(\omega\right)\in\left[0,1\right]$ for all $\omega\in\Omega$.
2. $\mathbb{1}_\Omega\in\left[\frac{d\mu}{d\mu}\right]_{\mathcal{A},\mu}$.

Proof Both claims follow from proposition 10. The first one requires the following result from Ash & Doléans-Dade. Probability and Measure Theory. 1999, 2nd edition:

Theorem 1.6.11. If $\mu$ is $\sigma$-finite on $\mathcal{F}$, $g$ and $h$ are Borel measurable, $\int_\Omega gd\mu$ and $\int_\Omega fd\mu$ exist, and $\int_A gd\mu\leq\int_a hd\mu$ for all $A\in\mathcal{F}$, then $g\leq h$ a.e. $\left[\mu\right]$.
$\square$
12. Proposition Let $S=\left(\Omega,\mathcal{A}\right)$ be a measurable space and let $\tau$, $\nu$ and $\mu$ be $\sigma$-finite measures on $S$.
1. If $\tau\ll\nu$ and $\nu\ll\mu$, then $\tau\ll\mu$ and $$\left[\frac{d\tau}{d\mu}\right]_{\mathcal{A},\mu} = \left[\frac{d\nu}{d\mu}\right]_{\mathcal{A},\nu}\left[\frac{d\nu}{d\mu}\right]_{\mathcal{A},\mu}$$
2. If $\tau,\nu\ll\mu$, $\tau+\nu\ll\mu$. If, additionally, $$\mathfrak{S}\left(\left[\frac{d\tau}{d\mu}\right]_{\mathcal{A},\mu},\left[\frac{d\nu}{d\mu}\right]_{\mathcal{A},\mu}\right)\neq\emptyset$$ then $$\left[\frac{d\left(\tau+\nu\right)}{d\mu}\right]_{\mathcal{A},\mu}=\left[\frac{d\tau}{d\mu}\right]_{\mathcal{A},\mu}+\left[\frac{d\nu}{d\mu}\right]_{\mathcal{A},\mu}$$
13. Proposition Let $S=\left(\Omega,\mathcal{A}\right)$ be a measurable space, let $\nu$ and $\mu$ be $\sigma$-finite measures on $S$ such that $\nu\ll\mu$ and let $f\in\left[\frac{d\nu}{d\mu}\right]_{\mathcal{A},\mu}$.

1. $\nu\left(\left\{\omega\in\Omega\mid:f\left(\omega\right)=0\right\}\right) = 0$.
2. For every $D\in\mathcal{A}$ such that $\nu(D^c)=0$ (where $D^c:=\Omega\backslash D$), $f\mathbb{1}_D\in\left[\frac{d\nu}{d\mu}\right]_{\mathcal{A},\mu}$.

Proof

1. Define $E:=\left\{\omega\in\Omega\mid:f\left(\omega\right)=0\right\}$. Then $$\nu(E) = \int_E\frac{d\nu}{d\mu}\space d\mu = \int_E f\space d\mu = \int_E 0\space d\mu = 0$$
2. For all $B\in\mathcal{A}$, $$\begin{array}{lcl} \int_B f\mathbb{1}_D d\mu & = & \int_{B\cap D}f d\mu=\int_{B\cap D}\frac{d\nu}{d\mu}d\mu \\ & = &\nu\left(B\cap D\right)=\nu(B)-\nu\left(B\cap D^c\right)=\nu(B) \end{array}$$ The result now follows from Theorem 10. $\square$

14. Proposition Let $S=\left(\Omega,\mathcal{A}\right)$ be a measurable space, let $\nu$, $\mu$ be measures on $S$ such that for all $B\in\mathcal{A}$, $\nu(B)\leq\mu(B)$, and let $f\in\left(\Omega,\mathcal{A}\right)\rightarrow\left(\overline{\mathbb{R}},\mathfrak{B}\right)$. If $0\leq f\left(\omega\right)$ for all $\omega\in\Omega$, $\int_\Omega f d\nu\leq\int_\Omega f d\mu$; and if $f\left(\omega\right)\leq0$ for all $\omega\in\Omega$, $\int_\Omega f d\mu\leq\int_\Omega f d\nu$.

Proof The case $0\leq f$ is proved first for indicator functions, then for non-negative simple functions and finally for the general case using the monotone convergence theorem.

Now, suppose $f\leq 0$. Then by the first case and by linearity of the integral, $$\int_\Omega f d\mu=-\int_\Omega -f d\mu\leq-\int_\Omega -f d\nu=\int_\Omega f d\nu$$

$\square$

## A Restatement of the Result to Be Proved

Let $S=\left(\Omega,\mathcal{A}\right)$ be a measurable space and let $\mu_0$, $\mu_1$, and $\mu_2$ be $\sigma$-finite measures on $S$. Suppose $$\begin{array}{lcl} f_1 & \in & \left[\frac{d\mu_0}{d\left(\mu_0+\mu_1\right)}\right]_{\mathcal{A},\mu_0+\mu_1} \\ f_2 & \in & \left[\frac{d\mu_0}{d\left(\mu_0+\mu_2\right)}\right]_{\mathcal{A},\mu_0+\mu_2} \end{array}$$

Then we have $T\left(f_1,f_2\right)\in\left[\frac{d\mu_0}{\mu_0+\mu_1+\mu_2}\right]_{\mu_0+\mu_1+\mu_2}$, where $T$ is an operator that takes two functions $f,g:\Omega\rightarrow\overline{\mathbb{R}}$ and returns the function $t:\Omega\rightarrow\overline{\mathbb{R}}$ defined thus: $$t\left(\omega\right) := \begin{cases} \frac{f\left(\omega\right)g\left(\omega\right)}{f\left(\omega\right)+g\left(\omega\right)-f\left(\omega\right)g\left(\omega\right)} &\mbox{if }f\left(\omega\right)+g\left(\omega\right)-f\left(\omega\right)g\left(\omega\right)\notin\left\{0,\pm\infty\right\},\\ 0 &\mbox{otherwise} \end{cases}$$

## Proof

The proof breaks into two parts:

1. Existence, where it is shown that there are certain $$\begin{array}{lcl} f_1 & \in & \left[\frac{d\mu_0}{d\left(\mu_0+\mu_1\right)}\right]_{\mathcal{A},\mu_0+\mu_1} \\ f_2 & \in & \left[\frac{d\mu_0}{d\left(\mu_0+\mu_2\right)}\right]_{\mathcal{A},\mu_0+\mu_2} \end{array}$$ for which $T\left(f_1,f_2\right)\in\left[\frac{d\mu_0}{\mu_0+\mu_1+\mu_2}\right]_{\mu_0+\mu_1+\mu_2}$.
2. Non-Uniqueness, where it is shown that for all $g_1\in\left[f_1\right]_{\mathcal{A},\mu_0+\mu_1}$ and $g_2\in\left[f_2\right]_{\mathcal{A},\mu_0+\mu_2}$, $T\left(g_1,g_2\right)\in\left[T\left(f_1,f_2\right)\right]_{\mathcal{A},\mu_0+\mu_1+\mu_2}$.

### Existence

Outline: After making some simplifying assumptions, we derive by gradual steps the equation $$\left(f_1+f_2-f_1f_2\right)\frac{d\mu_0}{d\mu}\mathbb{1}_D = f_1f_2\space\space\left[\mathcal{A},\mu\right]$$ for a certain set $D$. This equation is then seen to be equivalent to $T\left(f_1,f_2\right)\in\left[\frac{d\mu_0}{\mu_0+\mu_1+\mu_2}\right]_{\mu_0+\mu_1+\mu_2}$.

In light of proposition 11, we will assume w.l.g. that whenever $\mu$, $\nu$ are $\sigma$-finite measures on $S$ with $\nu(B)\leq\mu(B)$ for all $B\in\mathcal{A}$,

• $\frac{d\nu}{d\mu}\left(\omega\right)\in\left[0,1\right]$ for all $\omega\in\Omega$.
• $\frac{d\mu}{d\mu}=\mathbb{1}_\Omega$.
• Set $$\begin{array}{lcl} f_1 & := & \frac{d\mu_0}{d\left(\mu_0+\mu_1\right)} \\ f_2 & := & \frac{d\mu_0}{d\left(\mu_0+\mu_2\right)} \\ \mu & := & \mu_0+\mu_1+\mu_2 \end{array}$$ By proposition 12.1, $$\begin{array}{lcl} \frac{d\mu_0}{d\mu} & = & f_1\frac{d\left(\mu_0+\mu_1\right)}{d\mu}\space\space\left[\mathcal{A},\mu\right]\\ \frac{d\mu_0}{d\mu} & = & f_2\frac{d\left(\mu_0+\mu_2\right)}{d\mu}\space\space\left[\mathcal{A},\mu\right] \end{array}$$

By proposition 12.2, $$\begin{array}{lcl} \frac{d\mu_0}{d\mu} & = & f_1\left(\frac{d\mu_0}{d\mu}+\frac{d\mu_1}{d\mu}\right)\space\space\left[\mathcal{A},\mu\right]\\ \frac{d\mu_0}{d\mu} & = & f_2\left(\frac{d\mu_0}{d\mu}+\frac{d\mu_2}{d\mu}\right)\space\space\left[\mathcal{A},\mu\right] \end{array}$$

Multiply throughout by $f_2$ and $f_1$, respectively, and add up to obtain $$\left(f_1+f_2\right)\frac{d\mu_0}{d\mu} = 2f_1f_2\frac{d\mu_0}{d\mu} + f_1f_2\frac{d\mu_1}{d\mu} + f_1f_2\frac{d\mu_2}{d\mu}\space\space\left[\mathcal{A},\mu\right]$$

Assumption A allows us to subtract $f_1f_2\frac{d\mu_0}{d\mu}$ from both sides of the equation with a clear concience to obtain $$\left(f_1+f_2-f_1f_2\right)\frac{d\mu_0}{d\mu} = f_1f_2\frac{d\mu_0}{d\mu} + f_1f_2\frac{d\mu_1}{d\mu} + f_1f_2\frac{d\mu_2}{d\mu}\space\space\left[\mathcal{A},\mu\right]$$

A further application of proposition 12.2 yields $$\left(f_1+f_2-f_1f_2\right)\frac{d\mu_0}{d\mu} = f_1f_2\frac{d\mu}{d\mu}\space\space\left[\mathcal{A},\mu\right]$$

By assumption B, $$\left(f_1+f_2-f_1f_2\right)\frac{d\mu_0}{d\mu} = f_1f_2\space\space\left[\mathcal{A},\mu\right]$$

Define $$D:=\left\{\omega\in\Omega\mid:f_1\left(\omega\right)+f_2\left(\omega\right)-f_1\left(\omega\right)f_2\left(\omega\right)\neq0\right\}$$ If we can show that $\mu_0(D^c)=0$, we will have $$\left(f_1+f_2-f_1f_2\right)\frac{d\mu_0}{d\mu}\mathbb{1}_D = f_1f_2\space\space\left[\mathcal{A},\mu\right]$$ since by proposition 13.2, $\frac{d\mu_0}{d\mu}\mathbb{1}_D\in\left[\frac{d\mu_0}{d\mu}\right]_{\mathcal{A},\mu}$. This will amount to saying that $\frac{d\mu_0}{d\mu}\mathbb{1}_D = T\left(f_1,f_2\right)\space\space\left[\mathcal{A},\mu\right]$, since by assumption A, $f_1\left(\omega\right),f_2\left(\omega\right)\in\left[0,1\right]$ for all $\omega\in\Omega$. But then $T\left(f_1,f_2\right)\in\left[\frac{d\mu_0}{d\mu}\right]_{\mathcal{A},\mu}$, as was to be shown.

To see that $\mu_0\left(D^c\right)=0$, note that since $$\left\{\left(x_1,x_2\right)\in\left[0,1\right]\times\left[0,1\right]\mid:x_1+x_2-x_1x_2=0\right\}=\left\{\left(0,0\right)\right\}$$ it suffices to show that $\mu_0\left(\left\{\omega\in\Omega\mid:f_i\left(\omega\right)=0\right\}\right)=0$, for either $i=1$ or $i=2$. In fact, both cases hold, since by proposition 13.1, $$\mu_0\left(\left\{\omega\in\Omega\space\left|:\frac{d\mu_0}{d\left(\mu_0+\mu_i\right)}\left(\omega\right)=0\right.\right\}\right)=0,\space\space i=1,2$$

### Non-Uniqueness

Outline: We start by restating the problem in the form $\mu(E)=0$ for a certain set $E$. By considering subsets of $E$ and then subsets of these subsets, we are able to reduce the problem to that of showing that $\mu_2$ vanishes on a certain collection of subsets of $E$, $\left\{F_-,F_+, G_-, G_+\right\}$. The vanishing of $\mu_2$ on these sets is then shown to follow from proposition 7.

Let $g_1\in\left[f_1\right]_{\mathcal{A},\mu_0+\mu_1}$ and $g_2\in\left[f_2\right]_{\mathcal{A},\mu_0+\mu_2}$. Define $$\begin{array}{lcl} E & := & \left\{\omega\in\Omega\mid:T\left(g_1,g_2\right)\left(\omega\right)\neq T\left(f_1,f_2\right)\left(\omega\right)\right\} \\ \end{array}$$ It is our goal to show that $\mu(E)=0$.

Define $$\begin{array}{lcl} H_1 & := & \left\{\omega\in\Omega\mid:f_1\left(\omega\right)\neq g_1\left(\omega\right)\right\} \\ H_2 & := & \left\{\omega\in\Omega\mid:f_2\left(\omega\right)\neq g_2\left(\omega\right)\right\} \\ H & := & H_1\cup H_2 \end{array}$$ Since $E\subseteq H$, it suffices to show that $\mu(E\cap H_i)=0$, $i=1,2$. We only show that $\mu\left(E\cap H_1\right)=0$; the other equality can be deduced analogously.

Since, by definition of $g_1$, $\left(\mu_0+\mu_1\right)\left(H_1\right)=0$, we are left to show that $\mu_2\left(E\cap H_1\right)=0$.

Define $$\begin{array}{lcl} F_- & := & \left\{\omega\in E\cap H_1\mid:f_2\left(\omega\right)<0\right\} \\ F_+ & := & \left\{\omega\in E\cap H_1\mid:f_2\left(\omega\right)>0\right\} \\ F_\pm & := & F_-\cup F_+ \\ G_- & := & \left\{\omega\in E\cap H_1\mid:g_2\left(\omega\right)<0\right\} \\ G_+ & := & \left\{\omega\in E\cap H_1\mid:g_2\left(\omega\right)>0\right\} \\ G_\pm & := & G_-\cup G_+ \\ \end{array}$$ Since $E\cap H_1\subseteq F_\pm\cup G_\pm$, it suffices to show that $\mu_2\left(F_-\right)=\mu_2\left(F_+\right)=\mu_2\left(G_-\right)=\mu_2\left(G_+\right)=0$.

First we show that $\mu_2\left(F_-\right)=0$. Indeed, $$0\leq\mu_0\left(F_-\right)=\int_{F_-}\frac{d\mu_0}{d\left(\mu_0+\mu_2\right)}d\left(\mu_0+\mu_2\right)=\int_{F_-}f_2d\left(\mu_0+\mu_2\right)\leq\int_{F_-}f_2d\mu_2\leq0$$ where proposition 14 justifies the penultimate inequality. So $\int_{F_-}f_2d\mu_2=0$. Now by proposition 7, $\mu_2\left(F_-\right)=0$.

Next we show that $\mu_2\left(F_+\right)=0$. Indeed, $$\begin{array}{lcl} 0 & \leq & \int_{F_+}f_2d\mu_2\leq\int_{F_+}f_2d\left(\mu_0+\mu_2\right)=\int_{F_+}\frac{d\mu_0}{d\left(\mu_0+\mu_2\right)}d\left(\mu_0+\mu_2\right) \\ & = & \mu_0\left(F_+\right)\leq\left(\mu_0+\mu_1\right)\left(H_1\right)=0 \end{array}$$ where proposition 14 justifies the second inequality. So $\int_{F_+}f_2d\mu_2=0$. Invoking proposition 7 again, $\mu_2\left(F_+\right)=0$.

To show that $\mu_2\left(G_-\right)=\mu_2\left(G_+\right)=0$, employ the same argument, substituting $G_-$, $G_+$ for $F_-$, $F_+$, respectively. $\square$