Need an explanation of this paragraph "Measure Theory" I will just quote a part of one proof in "On uniformly regular topological measure spaces by Babiker: page 781" vol43 No4 Duke Math. J. 1976.

Let $I$ be the unit interval endowed with Lebesgue measure $m$. If $\mu$ is nonatomic probability (Radon) measure on (compact) space $X$, then there is $E\subset X$, $D\subset I$ with $\mu(E)=m(D)=0$ and a map $\psi\colon X\rightarrow I$ such that $\psi$ is a homeomorphism from $X\setminus E$ onto $I\setminus D$ and $m=\psi(\mu)$.

The author refer to this reference: N. Bourbaki, Elements De Mathematique Integration I, chaps 1-5, Paris, 1965: in chapter V page 85.
I do not know French to look at it. Could someone give me some steps proving the above result.
 A: This is what I found in my Russian edition of this book.
Lemma 1. Let $\mu$ be positive Borel measure on locally compact topological space $X$ and $A\subset X$ be $\mu$-measurable set with $\mu(A)>0$. Assume for all $\mu$-measurable $B\subset A$ either $\mu(B)=0$ or $\mu(B)=A$, then there exist $a\in A$ such that $\mu(A)=\mu(\{a\})$.
Idea. Consider intersection $P$ of all compact sets $K\subset A$ so that $\mu(K)=\mu(A)$. It will be the desired set.
Lemma 2. Let $\mu$ and $\nu$ be two positive Borel measures on $\mathbb{R}$. Assume that $\mu([a,b))=\nu([a,b))$ for all $a<b$. Then $\mu=\nu$.
Lemma 3. Let $X$ be compact topological space and $\mu$ be positive non-atomic probability measure on $X$. Then there continuous map $\pi:X\to I$ to the unit interval $I$. Such that image of $\mu$ under $\pi$ is Lebesgue measure $\lambda$ on $I$.
Sketch. Following idea of the proof of the Uryson's lemma we construct a family of open sets $U(t)$ where $t\in I$ such that $U(0)=\varnothing$, $U(1)=X$, $\overline{U(t)}\subset U(t')$ for $t'<t$ and finally $\mu(U(t))=t$ for all $t\in I$. Using lemma 1 we can show that conditions $V$, $W$ two open $\mu$-measurable with $\overline{V}\subset W$ and $\mu(V)<\mu(W)$ implies existence of $\mu$-meausrable open set $U$ such that $\overline{V}\subset U\subset\overline{U}\subset W$ such that $1/3\mu(W\setminus V)\leq\mu(U\setminus V)\leq 2/3\mu(W\setminus V)$. Now we apply lemma 2.
Theorem 1. Assume all assumptions of previous lemma are satisfied plus $X$ is metrizable, then there exist $\mu$-measurable set $N\subset X$ of measure $0$ and $\lambda$-measurable subset $M\subset I$ of measure zero such that we have a homeomorphism $\pi:I\setminus M \to X\setminus N$ with the following property once $\pi$ is continuously (in fact arbitrarily) extended up to $\varphi:I\to X$ and $\pi^{-1}$ similarly extended to $\psi:X\to I$, then $\varphi\circ\lambda=\mu$ and $\psi\circ\mu=\lambda$
Sketch. For each $n\in\mathbb{N}$ we can find finite partition of $X$ consisting of set of $\mu$-measure $0$ and open $\mu$-measurable sets of diameter $\leq1/n$ and measure $\leq 1/n$. (Don't ask why $-$ this is a long story) Then using induction and taking the limit we get existence of continuous map $f:I\setminus S\to X$ (where $S$ is countable) such that $f\circ\lambda=\mu$. Moreover $f$ can be chosen to be homeomorphism from $I\setminus S$ onto the subset of $X$ of full measure.
