Let $X$ be a set and give $X$ the discrete topology. Give $X\cup\{\infty\}$ the one-point compactification $X$. Then every subset $U$ of $X$ is open in $X\cup\{\infty\}$. Therefore, every subset of $U$ of $X\cup\{\infty\}$ is a Borel set. On the other hand, $X\cup\{\infty\}$ can be embedded as a closed subspace of $2^{\kappa}$ for some $\kappa$. Take note that the Borel subsets of $X\cup\{\infty\}$ are simply the restrictions of the Borel sets to $X\cup\{\infty\}$. Therefore, the power set $\sigma$-algebra $(X\cup\{\infty\},P(X\cup\{\infty\})$ can be embedded as Borel subset of $2^{\kappa}$.

A closer examination of my argument shows that a if $X$ can be embedded as a Borel set of some compactification (for example for locally compact spaces), then the $\sigma$-algebra $(X,Bor(X))$ (here $Bor(X)$ stands for the collection of all Borel subsets of $X$) can be embedded as a Borel subset of $2^{\kappa}$.

On the other hand, if $B\subseteq 2^{\kappa}$ is a Borel set, then the Borel measure on $B$ is the Borel measure of the topology on $B.$ However, by this answer not every $\sigma$-algebra is the Borel measure of some topology.

Let $X$ be a set and give $X$ the discrete topology. Give $X\cup\{\infty\}$ the one-point compactification $X$. Then every subset $U$ of $X$ is open in $X\cup\{\infty\}$. Therefore, every subset of $U$ of $X\cup\{\infty\}$ is a Borel set. On the other hand, $X\cup\{\infty\}$ can be embedded as a closed subspace of $2^{\kappa}$ for some $\kappa$. Take note that the Borel subsets of $X\cup\{\infty\}$ are simply the restrictions of the Borel sets to $X\cup\{\infty\}$. Therefore, the power set $\sigma$-algebra $(X\cup\{\infty\},P(X\cup\{\infty\})$ can be embedded as Borel subset of $2^{\kappa}$.

A closer examination of my argument shows that a if $X$ can be embedded as a Borel set of some compactification (for example for locally compact spaces), then the $\sigma$-algebra $(X,Bor(X))$ (here $Bor(X)$ stands for the collection of all Borel subsets of $X$) can be embedded as a Borel subset of $2^{\kappa}$.

On the other hand, if $B\subseteq 2^{\kappa}$ is a Borel set, then the Borel measure on $B$ is the Borel measure of the topology on $B.$ However, by this answer not every $\sigma$-algebra is the Borel measure of some topology.

Let $X$ be a set and give $X$ the discrete topology. Give $X\cup\{\infty\}$ the one-point compactification $X$. Then every subset $U$ of $X$ is open in $X\cup\{\infty\}$. Therefore, every subset of $U$ of $X\cup\{\infty\}$ is a Borel set. On the other hand, $X\cup\{\infty\}$ can be embedded as a closed subspace of $2^{\kappa}$ for some $\kappa$. Take note that the Borel subsets of $X\cup\{\infty\}$ are simply the restrictions of the Borel sets to $X\cup\{\infty\}$. Therefore, the power set $\sigma$-algebra $(X\cup\{\infty\},P(X\cup\{\infty\})$ can be embedded as Borel subset of $X$.

On the other hand, if $B\subseteq 2^{\kappa}$ is a Borel set, then the Borel measure on $B$ is the Borel measure of the topology on $B.$ However, by this answer not every $\sigma$-algebra is the Borel measure of some topology.

Let $X$ be a set and give $X$ the discrete topology. Give $X\cup\{\infty\}$ the one-point compactification $X$. Then every subset $U$ of $X$ is open in $X\cup\{\infty\}$. Therefore, every subset of $U$ of $X\cup\{\infty\}$ is a Borel set. On the other hand, $X\cup\{\infty\}$ can be embedded as a closed subspace of $2^{\kappa}$ for some $\kappa$. Take note that the Borel subsets of $X\cup\{\infty\}$ are simply the restrictions of the Borel sets to $X\cup\{\infty\}$. Therefore, the power set $\sigma$-algebra $(X\cup\{\infty\},P(X\cup\{\infty\})$ can be embedded as Borel subset of $2^{\kappa}$.

A closer examination of my argument shows that a if $X$ can be embedded as a Borel set of some compactification (for example for locally compact spaces), then the $\sigma$-algebra $(X,Bor(X))$ (here $Bor(X)$ stands for the collection of all Borel subsets of $X$) can be embedded as a Borel subset of $2^{\kappa}$.

On the other hand, if $B\subseteq 2^{\kappa}$ is a Borel set, then the Borel measure on $B$ is the Borel measure of the topology on $B.$ However, by this answer not every $\sigma$-algebra is the Borel measure of some topology.