If no outer measure is present, do people use the term "measurable set" simply for any element in a given $\sigma$-algebra?

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This is a really silly question, but I just need to be certain about this.

Let $(X,\mathcal{A})$ be a measurable space. Often authors talk about measurable sets, even when we're not in the presence of an outer measure. So, when they talk about "measurable sets", they simply mean any set in $\mathcal{A}$, right?

So, for example, if $\mu$ is a signed measure on $(X,\mathcal{A})$, then the definition "$A$ is a positive set with respect to $\mu$ if $A$ is measurable and for every measurable subset $E$ of $A$ we have $\mu(E) \geq 0$" can be translated into "$A$ is a positive set with respect to $\mu$ if $A \in \mathcal{A}$ and for every measurable subset $E \subseteq A$ such that $E \in \mathcal{A}$ we have $\mu(E) \geq 0$", right?

Thanks.

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As I have been invited to answer this question: Yes, that's correct. As a reference for instance Rudin's 'Real and Complex Analysis' (chap. I, the concept of measurability): if $\mathfrak{M}$ is a $\sigma$-algebra on $X$, then $(X,\mathfrak{M})$ is said to be a measurable space and the members of $\mathfrak{M}$ are said to be the measurable sets in $X$. (Notice that several definitions are formally quite similar as in topology.)