Is nonsingular group action on a measure space "continuous"?

Suppose $G$ is a group acting on a nonatomic standard measure space $(X,\mu)$ (say $[0,1]$ with Lebesgue measure). Assume that the action is nonsingular, i.e. $\mu(E)=0$ implies $\mu(gE)=0$ for all $g\in G$ and measurable subsets $E\subset X$.

Given $g_0\in G$ and $\delta>0$, is it true that there exists $\varepsilon>0$ such that $\mu(E)\leq\varepsilon$ implies $\mu(g_0E)\leq\delta$ for all measurable $E\subset X$?

This seems to make sense, but I don't immediately see how to start (dis)proving it.

• Are you familiar with absolute continuity of measures? For each $g \in G$, we can define a measure $\mu_{g}$ by $\mu_{g}(E) = \mu(gE)$. (I leave it to you to show this defines a measure.) Your definition of nonsingular action says that $\mu_{g}$ is an absolutely continuous measure (for each $g$). Mar 23 '16 at 3:31

As $g_0$ is fixed we need only consider an invertible non-singular transformation $T : X \rightarrow X$.
Since the measures $\mu$ and $\mu \circ T$ are equivalent by the Radon-Nikodym theorem we have an integrable function $f > 0$ almost everywhere such that $$\mu(TE) = \mu \circ T(E) = \int_E f \, d\mu$$ for every measurable set $E$.
For any $\eta = \delta/2 > 0$ we can find a simple function $s(x)$ such that $f \geq s$ on $X$ and $$\int_X f - s \, d\mu < \eta.$$ Since $s$ is bounded, by $M >0$ say, we have that $$\mu(TE) = \int_E f - s \, d\mu + \int_E s \, d\mu \leq \eta + M\mu(E).$$ So taking $\mu(E) < \eta/M$ suffices to prove the claim.