[proofplan]
We quotient positive-measure Borel sets by equality modulo null sets and compare the resulting order with the Boolean order in the measure algebra. The map sending a Borel set to its equivalence class is first shown to be well-defined on the quotient forcing. We then prove that it preserves and reflects the order, preserves compatibility through positive-measure intersections, and is onto. Therefore it is an order isomorphism, hence a dense embedding; the standard dense-embedding argument for forcing gives the asserted equivalence of forcing notions.
[/proofplan]
[step:Pass the class map through the null-symmetric-difference quotient]
Let $\mu$ denote the standard product probability measure on $2^\omega$. Let $\mathbb B_{\mathrm{rand}}$ denote the Boolean algebra of Borel subsets of $2^\omega$ modulo the ideal of $\mu$-null Borel sets, and let $\mathbb B_{\mathrm{rand}}^+$ denote its nonzero part. Define
\begin{align*}
q:P_{\mathrm{Borel}}\to \mathbb B_{\mathrm{rand}}^+
\end{align*}
by $q(A)=[A]$, where $[A]$ denotes the equivalence class of the Borel set $A$ modulo the ideal of $\mu$-null Borel sets. Since $A\in P_{\mathrm{Borel}}$ means $\mu(A)>0$, the class $[A]$ is nonzero in $\mathbb B_{\mathrm{rand}}$, so $q(A)\in \mathbb B_{\mathrm{rand}}^+$.
If $A,C\in P_{\mathrm{Borel}}$ and $A\sim C$, then $\mu(A\triangle C)=0$, which is precisely the defining [equivalence relation](/page/Equivalence%20Relation) of the measure algebra. Hence $[A]=[C]$. Therefore $q$ is constant on $\sim$-equivalence classes and induces a map
\begin{align*}
\bar q:P_{\mathrm{Borel}}/{\sim}\to \mathbb B_{\mathrm{rand}}^+
\end{align*}
defined by $\bar q([A]_{\sim})=[A]$, where $[A]_{\sim}$ denotes the class of $A$ in the quotient presentation.
[guided]
The first issue is that $q$ is initially defined on actual Borel sets, while the quotient forcing identifies two conditions whenever they differ only on a null set. We must check that the value of $q(A)$ does not depend on which representative of the quotient class is chosen.
Let $\mu$ denote the standard product probability measure on $2^\omega$. Let $\mathbb B_{\mathrm{rand}}$ denote the Boolean algebra of Borel subsets of $2^\omega$ modulo the ideal of $\mu$-null Borel sets, and let $\mathbb B_{\mathrm{rand}}^+$ denote its nonzero part. Let
\begin{align*}
q:P_{\mathrm{Borel}}\to \mathbb B_{\mathrm{rand}}^+
\end{align*}
be the map $q(A)=[A]$, where $[A]$ is the measure-algebra class of the Borel set $A$. This lands in $\mathbb B_{\mathrm{rand}}^+$ because $A\in P_{\mathrm{Borel}}$ implies $\mu(A)>0$, so $A$ is not null and hence $[A]\ne 0$ in the measure algebra.
Now suppose $A,C\in P_{\mathrm{Borel}}$ satisfy $A\sim C$. By definition of $\sim$,
\begin{align*}
\mu(A\triangle C)=0.
\end{align*}
But the measure algebra identifies exactly those Borel sets whose symmetric difference has measure zero. Therefore $[A]=[C]$ in $\mathbb B_{\mathrm{rand}}$. Thus the assignment $A\mapsto [A]$ is constant on each $\sim$-equivalence class.
Consequently there is a well-defined induced map
\begin{align*}
\bar q:P_{\mathrm{Borel}}/{\sim}\to \mathbb B_{\mathrm{rand}}^+
\end{align*}
given by $\bar q([A]_{\sim})=[A]$. The notation $[A]_{\sim}$ is used for the quotient class in $P_{\mathrm{Borel}}/{\sim}$, while $[A]$ is the measure-algebra class.
[/guided]
[/step]
[step:Show that the induced map preserves and reflects the order]
Let $[A]_{\sim},[C]_{\sim}\in P_{\mathrm{Borel}}/{\sim}$. The quotient forcing order is the order induced by inclusion modulo null sets: $[A]_{\sim}\le [C]_{\sim}$ means that some, equivalently every, representative of $[A]_{\sim}$ is contained in a representative of $[C]_{\sim}$ modulo a null set. Equivalently,
\begin{align*}
[A]_{\sim}\le [C]_{\sim} \iff \mu(A\setminus C)=0.
\end{align*}
This condition is independent of representatives: if $\mu(A\triangle A')=0$ and $\mu(C\triangle C')=0$, then $A'\setminus C'\subset (A'\triangle A)\cup(A\setminus C)\cup(C\triangle C')$, so $\mu(A'\setminus C')=0$ whenever $\mu(A\setminus C)=0$.
The Boolean order on $\mathbb B_{\mathrm{rand}}^+$ is also defined by
\begin{align*}
[A]\le [C] \iff \mu(A\setminus C)=0.
\end{align*}
Therefore
\begin{align*}
[A]_{\sim}\le [C]_{\sim} \iff \bar q([A]_{\sim})\le \bar q([C]_{\sim}).
\end{align*}
Thus $\bar q$ preserves and reflects the forcing order.
[/step]
[step:Identify compatibility with positive-measure intersection]
Let $[A]_{\sim},[C]_{\sim}\in P_{\mathrm{Borel}}/{\sim}$. They are compatible in the quotient forcing exactly when there exists $D\in P_{\mathrm{Borel}}$ such that $[D]_{\sim}\le [A]_{\sim}$ and $[D]_{\sim}\le [C]_{\sim}$.
If such a $D$ exists, then $\mu(D)>0$, $\mu(D\setminus A)=0$, and $\mu(D\setminus C)=0$. Hence
\begin{align*}
\mu(D\setminus (A\cap C))=0.
\end{align*}
Since $\mu(D)>0$, it follows that $\mu(A\cap C)>0$.
Conversely, if $\mu(A\cap C)>0$, define the Borel set
\begin{align*}
D:=A\cap C.
\end{align*}
Then $D\in P_{\mathrm{Borel}}$, and $D\subset A$ and $D\subset C$, so $[D]_{\sim}\le [A]_{\sim}$ and $[D]_{\sim}\le [C]_{\sim}$. Hence $[A]_{\sim}$ and $[C]_{\sim}$ are compatible.
In the measure algebra, the meet of $[A]$ and $[C]$ is $[A\cap C]$, and this meet is nonzero exactly when $\mu(A\cap C)>0$. Therefore compatibility in the quotient presentation agrees with compatibility in $\mathbb B_{\mathrm{rand}}^+$.
[/step]
[step:Prove surjectivity and conclude dense embedding]
Let $b\in \mathbb B_{\mathrm{rand}}^+$. By definition of the measure algebra, there is a Borel set $A\subset 2^\omega$ such that $b=[A]$. Since $b\ne 0$, the set $A$ is not null, so $\mu(A)>0$. Thus $A\in P_{\mathrm{Borel}}$, and
\begin{align*}
\bar q([A]_{\sim})=[A]=b.
\end{align*}
Therefore $\bar q$ is surjective. Since it also preserves and reflects the order, $\bar q$ is an order isomorphism from $P_{\mathrm{Borel}}/{\sim}$ onto $\mathbb B_{\mathrm{rand}}^+$.
An order isomorphism is, in particular, a dense embedding because its image is all of the target forcing. The standard dense-embedding argument for forcing says that if $e:P\to Q$ is a dense embedding of forcing posets, then $P$-generic filters correspond to $Q$-generic filters by taking the generated filter through $e$, and the resulting generic extensions are the same. In the present case $e=\bar q$ is stronger than a dense embedding because it is onto and order-reflecting. Hence a filter $G\subset P_{\mathrm{Borel}}/{\sim}$ is generic exactly when $\bar q[G]\subset \mathbb B_{\mathrm{rand}}^+$ is generic, and the inverse image under $\bar q$ recovers $G$. Therefore the two forcing notions produce the same generic extensions up to the canonical relabelling of conditions by $\bar q$. This proves that the measure algebra presents random forcing.
[/step]
