In Bogachev's Measure Theory, he states the following:
Suppose we are given two spaces $X,Y$ with $\sigma$-algebras $\mathcal{A}$ and $\mathcal{B}$ and an $(\mathcal{A},\mathcal{B})$-measurable mapping $f\colon X \to Y$. Then, for any bounded (or bounded from below) measure $\mu$ on $\mathcal{A}$, the formula \begin{equation} \mu \circ f^{-1} \colon B \mapsto \mu(f^{-1}(B)), \quad B \in \mathcal{B}. \end{equation}
In connection with this, he states the following result.
3.6.1 Theorem: Let $\mu$ be a nonnegative measure. A $\mathcal{B}$-measurable function $g$ on $Y$ is integrable with respect to the measure $\mu\circ f^{-1}$ precisely when the function $g\circ f$ is integrable with respect to $\mu$. In addtion, one has \begin{equation} \int_Y g(y) \mu\circ f^{-1}(dy) = \int_X g(f(x)) \mu(dx). \end{equation}
He defines a measurable function as
2.1.1 Definition: Let $(X,\mathcal{A})$ be a measurable space, i.e., a space with a $\sigma$-algebra. A function $g \colon X \to \mathbb{R}^1$ is called measurable with respect to $\mathcal{A}$ (or $\mathcal{A}$-measurable) if $\{x : g(x) < c\} \in \mathcal{A}$ for every $c \in \mathbb{R}^1$.
Does the same (or a similar) result hold if $g \colon X \to \mathbb{R}^n$, for arbitrary $n \in \mathbb{N}$? If not, what if we impose some extra conditions on $g$, like continuity and boundedness?