Is Gaussian (mean unknown) dominated by a $\sigma$-finite measure?

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I am reading about the existence of minimal sufficient statistic. Jun Shao has that

Minimal sufficient statistic exist under weak assumptions, e.g., $\mathcal{P}$ contains distributions on $\mathbb{R}^k$ dominated by a $\sigma$-finite measure.

I know the definition of a measure $\mu$ being dominated by another measure $v$ (i.e. $v\left(A\right) = 0$ implies $\mu\left(A\right) = 0$). But I would like to consider some more familiar examples, to understand why this is a "weak assumption".

For example, consider the parametric family of Gaussian density, with unknown mean.

$$P_\theta=\left\{\frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{(x-\mu)^2}{2}\right)\mid \mu \in \mathbb{R} \right\}$$

Is this family dominated by a $\sigma$-finite measure (defined on $\mathbb{R}$)? Is it just dominated by the Lebesgue measure on $\mathbb{R}$?

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Any probability measure having a density function is dominated by Lebesgue measure. So all Gaussian measures are dominated by Lebesgue measure.