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Let $\Omega$ be an infinite set and suppose that ${\cal F}$ is a $\sigma$-algebra on $\Omega$ such that $|{\cal F}|\geq 2^{\aleph_0}$.

Is there a measure $\mu : {\cal F} \to [0,1]$ such that $\mu$ is surjective as a function?

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$\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$

As Gerald Edgar noted, the answer is yes. Indeed, let $S:=\Omega$. Let us say that a set $B$ is a proper subset of a set $C$ if $B\subseteq C$, $B\ne\emptyset$, and $B\ne C$. Let us say that a subset $A$ of $S$ is an atom if $A\in\F$, $A\ne\emptyset$, and $A$ has no proper subset $B\in\F$. Note that any two atoms are either the same or disjoint.

One of the following cases must occur.

Case 1: there are only finitely many atoms and their union is $S$. Then $\F$ is finite, which contradicts the condition $|{\F}|\geq 2^{\aleph_0}$.

Case 2: there are only finitely many atoms and their union (say $U$) is not $S$. Let then $V_0:=S\setminus U$, so that $V_0$ has a proper subset $V_1\in\F$, which has a proper subset $V_2\in\F$, etc. Then the sets $W_j:=V_j\setminus V_{j-1}$ for $j=1,2,\dots$ are nonempty disjoint members of $\F$.

Case 3: there are infinitely many atoms $W_1,W_2,\dots$.

So, in both Cases 2 and 3, there are infinitely many nonempty disjoint members $W_1,W_2,\dots$ of $\F$. For each $i=1,2,\dots$, take any $w_i\in W_i$, and for each $B\in\F$ let \begin{equation} \mu(B):=\sum_1^\infty\frac1{2^i}\ii{w_i\in B}, \end{equation} where $\ii{\cdot}$ denotes the indicator.

Then $\mu$ is a surjective probability measure. Indeed, take any $p\in[0,1]$ and consider its binary expansion \begin{equation} p=\sum_1^\infty\frac{\de_i}{2^i} \end{equation} for some $\de_i\in\{0,1\}$. Let $J$ be the set of all natural numbers $i$ such $\de_i=1$. Then \begin{equation} \mu\Big(\bigcup_{i\in J}W_i\Big)=\sum_{i\in J}\mu(W_i)=\sum_{i\in J}\frac1{2^i}=p. \end{equation}

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