What is $F_\mu(\mu)$? Given a measure $\mu$ and the corresponding distribution function $F_\mu$.
What happens if one looks at
$$F_\mu(\mu)~?
$$
One might as well assume that $\mu <<\lambda$.
Thanks in advance!
 A: Distribution functions are really only defined for probability measures on the real line. That is,
$$
F_\mu(x)= \mu((-\infty,x]),
$$
in which case the distribution function is a map $F_\mu:\mathbb{R} \to [0,1]$. Here we note that the domain is the real numbers and as such is does not make any sense to evaluate the function in a probability measure $\mu$. 
If your are instead suggesting we look at the push-forward measure of $F_\mu$ on the Borel probability measure $\mu$, then its another scenario and it makes perfect sense. In this case what you are looking for is probably something like the following:  Proposition 3.1 in this pdf. If we insist on avoiding talking about random variables, we can replicate the arguments from the proposition in the following manner:
In your notation and using proposition from the above link, it would become something like this:
Note that for any set $(-\infty,a]$ we have that
\begin{align*}
F_\mu(\mu)((-\infty,a]) :&= \mu(F_\mu^{-1}((-\infty,a]))) \\
&= \mu( \{x\in \mathbb{R}: F_\mu(x)\in (-\infty,a]\}) \\
&=\mu( \{x\in \mathbb{R}: F_\mu(x)\leq a\}) \\
&= \left\{ \begin{array}{ll} \mu(\emptyset)=0, & \text{if } a<0 \\
\mu( \{x\in \mathbb{R}: F_\mu(x)\leq a\}), & \text{if } 0 \leq a \leq 1\\
\mu(\mathbb{R})=1,  &  \text{if } a>1\end{array} \right.,
\end{align*}
Let $F_\mu^{\leftarrow}$ denote generalized inverse of $F_\mu$ given by
$$
F_\mu^{\leftarrow}(y) =\inf\{ x\in\mathbb{R}: F_\mu(x)\geq y \}.
$$
Now assume that $F_\mu$ is continuous and let $0\leq a \leq 1$. First realize that the generalized inverse is strictly increasing on $\text{range}(F_\mu)=[0,1]$ (c.f. prop. 2.3 (7)) implying that
\begin{align*}
\mu( \{ x \in \mathbb{R}: F_\mu(x)\leq a\}) &=\mu( \{ x \in \mathbb{R}: F_\mu^{\leftarrow}(F_\mu(x))\leq F_\mu^{\leftarrow}(a\}) \\& =\mu( \{ x \in \text{supp}(\mu): F_\mu^{\leftarrow}(F_\mu(x))\leq F_\mu^{\leftarrow}(a\})
\end{align*}
where we in the last equality used that the support of $\mu$ has measure 1 (All Borel probability measures on a separable metric space has support of full measure). The support is formally defined by
$
\text{supp}(\mu) = \{x\in\mathbb{R}| \forall N\in \mathcal{N}_x: \mu(N)>0\},
$
where $\mathcal{N}_x$ is all open neighbourhoods of $x$. 
Finally we have to realize that $F_\mu$ is strictly increasing on $\text{supp}(\mu)$ such that prop. 2.3. (3) yields that
\begin{align*}
\mu( \{ x \in \text{supp}(\mu): F_\mu^{\leftarrow}(F_\mu(x))\leq F_\mu^{\leftarrow}(a\}) &= \mu( \{ x \in \text{supp}(\mu): x\leq F_\mu^{\leftarrow}(a\}) \\&= \mu( \{ x \in \mathbb{R}: x\leq F_\mu^{\leftarrow}(a\}) \\&= F_\mu(F_\mu^{\leftarrow}(a)) \\&= a
\end{align*}
where we in the last equality used prop 2.3 (4). 
Since sets of the form $(-\infty,a]$ is an intersection stable generator of $\mathcal{B}(\mathbb{R})$ and our measure $F_\mu(\mu)$ coincides with the uniform measure on $[0,1]$ on these sets, they must be equal.
