The following provides a necessary and sufficient condition for an arbitrary crossed product von Neumann algebra $L^\infty(X) \rtimes G$ to be a factor. As a corollary, I include a simpler criterion for actions that preserve a probability measure.

First note that a criterion for arbitrary actions necessarily has to refer to induced actions. Given an action $G \curvearrowright (X,\mu)$ of a countable group $G$ by nonsingular transformations (i.e. preserving sets of measure zero) of a standard probability space $(X,\mu)$, it may happen that there exists a subgroup $\Gamma < G$ and a $\Gamma$-invariant Borel set $Y \subset X$ such that, up to measure zero, the sets $(g \cdot Y)_{g \in G/\Gamma}$ form a partition of $X$. Then the crossed products satisfy
$$L^\infty(X) \rtimes G \cong B(\ell^2(G/\Gamma)) \overline{\otimes} (L^\infty(Y) \rtimes \Gamma) \; .$$
In particular, the crossed product for $G \curvearrowright X$ is a factor if and only if the crossed product for $\Gamma \curvearrowright Y$ is a factor.

Conversely, given any nonsingular action $\Gamma \curvearrowright Y$ and any embedding of $\Gamma$ into a larger group $G$, there is a natural action $G \curvearrowright X = G/\Gamma \times Y$ that is induced from $\Gamma \curvearrowright Y$.

These induced actions are characterized by the existence of a $G$-equivariant map $X \rightarrow G/\Gamma$. Note that if $\mu$ is an invariant probability measure, this forces $\Gamma$ to be a finite index subgroup of $G$; see the corollary below.

**Theorem.** Let $G$ be a countable group and $G \curvearrowright (X,\mu)$ any action by nonsingular transformations of a standard probability space $(X,\mu)$. Then the following two conditions are equivalent.

(1) The crossed product $L^\infty(X) \rtimes G$ is a factor.

(2) The action $G \curvearrowright (X,\mu)$ is ergodic and whenever $G \curvearrowright X$ is induced from $\Gamma \curvearrowright Y$, the following holds: if $h \in \Gamma \setminus \{e\}$ acts trivially on $Y$, then $h$ has an infinite $\Gamma$-conjugacy class.

**Proof.** Write $A = L^\infty(X)$ and $M = A \rtimes G$. View $A \subset M$ and denote by $(u_g)_{g \in G}$ the canonical unitary operators in $M$.

(1) $\Rightarrow$ (2). If $F \in A$ is $G$-invariant, then $F$ belongs to the center of $M$ and hence $F$ is constant. So $G \curvearrowright (X,\mu)$ is ergodic. Assume that $G \curvearrowright (X,\mu)$ is induced from $\Gamma \curvearrowright Y$. Denote by $N \lhd \Gamma$ the normal subgroup that acts trivially on $Y$ and take $h \in N \setminus \{e\}$. If the $\Gamma$-conjugacy class $C$ of $h$ is finite, then $\sum_{g \in C} u_g$ defines a non scalar central element of the crossed product $L^\infty(Y) \rtimes \Gamma$. Then also $M$ would fail to be a factor. So $C$ must be infinite.

$\neg$ (1) $\Rightarrow$ $\neg$ (2). Assume that $M$ is not a factor and that $G \curvearrowright (X,\mu)$ is ergodic. We have to prove that $G \curvearrowright X$ is induced from $\Gamma \curvearrowright Y$ in such a way that there exists an element $h \in \Gamma \setminus \{e\}$ that acts trivially on $Y$ and that has a finite $\Gamma$-conjugacy class.

Let $T$ be a non scalar element in the center of $M$. Denote by $E : M \rightarrow A$ the normal conditional expectation satisfying $E(u_g) = 0$ for all $g \in G \setminus \{e\}$. Then, $E(T)$ is $G$-invariant and thus scalar. Replacing $T$ by $T - E(T)$, we may thus assume that $T$ is a central element with $E(T) = 0$ and $T \neq 0$.

View $M$ as the algebra of bounded measurable functions $F : X \rightarrow B(\ell^2(G))$ satisfying $T(g \cdot x) = \rho_g T(x) \rho_g^*$ for all $g \in G$ and a.e. $x \in X$. Here $\rho$ is the right regular representation of $G$. We then associate to $T$ the function
$$f : X \rightarrow \ell^2(G) : f(x) = T(x) \delta_e \; .$$
Since $T(x) \delta_h = \rho_h f(h^{-1} \cdot x)$, we get that $f$ is not a.e. zero. Since $E(T) = 0$, we have $f(x) \in \ell^2(G \setminus \{e\})$. The fact that $T$ is central translates to the following two properties for the function $f$.

The fact that $T$ commutes with the unitary operators $u_g$ translates to: $f(g \cdot x) = \text{Ad}(g) f(x)$ for all $g \in G$ and a.e. $x \in X$.

The fact that $T$ commutes with $A$ translates to: if $f(x)(h) \neq 0$, then $h \cdot x = x$.

Define the measurable function
$$S : X \rightarrow \mathbb{R} : S(x) = \max \{ |f(x)(h)| \mid h \in G \} \; .$$
This is well defined because $f(x) \in \ell^2(G)$. Since $S$ is $G$-invariant, it follows that $S$ is constant a.e. Also, $S$ is not a.e. zero. So we find $0 < s < \infty$ such that $S(x) =s$ for a.e. $x \in X$. For every $x \in X$, denote by $R(x) \subset G$ the finite subset given by
$$R(x) = \{h \in G \mid |f(x)(h)| = s\} \; .$$
We have $R(g \cdot x) = g R(x) g^{-1}$. Also, $R(x) \neq \emptyset$ and $R(x) \subset G \setminus \{e\}$ for a.e. $x \in X$. There are only countably many finite subsets of $G$. So we can fix a nonempty finite subset $C \subset G \setminus \{e\}$ such that the set
$$Y = \{x \in X \mid R(x) = C\}$$
has positive measure. When $g \in G$, the set $g \cdot Y \cap Y$ has positive measure if and only if $g C g^{-1} = C$. Defining the subgroup $\Gamma < G$ by
$$\Gamma = \{g \in G \mid g C g^{-1} = C\} \; ,$$
it follows that $G \curvearrowright X$ is induced from $\Gamma \curvearrowright Y$. When $g \in C$ and $x \in Y$, we have $|f(x)(g)| = s > 0$ so that $g \cdot x = x$. In particular, $R(g \cdot x) = R(x)$, so that $gCg^{-1} = C$ and $g \in \Gamma$. We conclude that $C$ is a finite subset of $\Gamma \setminus \{e\}$ that is invariant under $\Gamma$-conjugation and that all elements of $C$ act trivially on $Y$. So, $\neg 2$ holds.

**Corollary.** Let $G$ be a countable group and $G \curvearrowright (X,\mu)$ any action by $\mu$-preserving transformations of a standard probability space $(X,\mu)$. Then the following two conditions are equivalent.

(1) The crossed product $L^\infty(X) \rtimes G$ is a factor.

(2) The action $G \curvearrowright (X,\mu)$ is ergodic and if $g \in G \setminus \{e\}$ is such that $\{x \in X \mid g \cdot x = x\}$ has positive measure, then $g$ has an infinite conjugacy class.

**Proof.** (2) $\Rightarrow$ (1). Assume that $G \curvearrowright X$ is induced from $\Gamma \curvearrowright Y$ and that $g \in \Gamma \setminus \{e\}$ acts trivially on $Y$. By the theorem above, it suffices to prove that $g$ has an infinite $\Gamma$-conjugacy class. Since $\mu$ is a $G$-invariant probability measure, $\Gamma < G$ has finite index. It thus suffices to prove that $g$ has an infinite $G$-conjugacy class. Since $g$ fixes all points of $Y$, this is indeed the case by the assumption in (2).

$\neg$ (2) $\Rightarrow$ $\neg$ (1). Assume that $g_0 \in G \setminus \{e\}$ fixes all points in a subset of $X$ of positive measure and that the $G$-conjugacy class $C$ of $g_0$ is finite. We prove that the second condition in the theorem above does not hold. Since $C$ is finite and since $g_0 \in C$ fixes all points in a set of positive measure, we can choose a maximal subset $C_0 \subset C$ containing $g_0$ and having the property that
$$Y := \{x \in X \mid h \cdot x = x \;\;\text{for all $h \in C_0$}\;\}$$
has positive measure. When $g \in G$ and $x \in g \cdot Y \cap Y$, we have that $h \cdot x = x$ for all $h \in C_0 \cup g C_0 g^{-1} \subset C$. By the maximality of $C_0$ we conclude that $g \cdot Y \cap Y$ has positive measure if and only if $gC_0 g^{-1} = C_0$ if and only if $g \cdot Y = Y$. Defining
$$\Gamma = \{g \in G \mid gC_0 g^{-1} = C_0\} \; ,$$
we find that $G \curvearrowright X$ is induced from $\Gamma \curvearrowright Y$. By construction, $C_0 \subset \Gamma \setminus \{e\}$ is a finite set of elements that act trivially on $Y$ and $C_0$ is invariant under conjugation by $\Gamma$. So the second condition in the theorem above does not hold and the corollary is proven.