If $(Ω,A,\text P)$ is a non-atomic prob. space and $μ$ is a prob. measure on a Borel space there is a variable on $(Ω,A,\text P)$ with law $μ$

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a non-atomic probability space
• $(E,\mathcal E)$ be a Borel space
• $\mu$ be a probability measure on $(E,\mathcal E)$

How can we show that there is a $(\mathcal A,\mathcal E)$-measurable $X:\Omega\to E$ with $X_\ast\operatorname P=\mu$ (pushforward measure)?

I know that if $(\mathbb R,\mathcal B(\mathbb R),\nu)$ is a non-atomic probability space, the distribution function $F$ of $\nu$ is continous and $F_\ast\nu=\mathcal U_{[0,\:1]}$ (uniform distribution).

Now, I'll assume that for all $B\in\mathcal B(\mathbb R)$, $(B,\mathcal B(B))$ is isomorphic to $(\mathbb R,\mathcal B(\mathbb R))$ (i.e. there is a measurable bijection between $(B,\mathcal B(B))$ and $(\mathbb R,\mathcal B(\mathbb R))$ with measurable inverse). Does anyone have a reference for that claim?

By that assumption, there is a $\mathcal U_{[0,\:1]}$-distributed random variable on any non-atomic probability space $(B,\mathcal B(B),\nu)$ with $B\in\mathcal B(\mathbb R)$.

$(E,\mathcal E)$ being Borel implies that $(E,\mathcal E)$ is isomorphic to $(B,\mathcal B(B))$.

How can we conclude? This answer seems to claim that the desired conclusion is possible, but I don't get how we need to argue exactly.