$\def\R{{\mathbb R}}$ Can I please receive help proving this? Thank you.
Suppose $(X,\mathcal{M}, \mu)$ is a complete measure space and $f: X\to \R$ is a measurable function. Let $\mu_f(B) = \mu(f^{-1}(B)).$ Show $\mu_f$ is a measure on the $\sigma-$algebra $\mathcal{B}$ of Borel subsets of $\R.$ Thus, $f$ induces a Borel measure $\mu_f$ on $\R.$
$\textbf{My work so far.}$ Observe that $\mu$ is a $\sigma$-finite, $\mu$ is saturated. Moreover, $\mathcal{B}$ is a $\sigma-$algebra. So, $\mu_f(B) = \mu(f^{-1}(B)).$ Thus, it is a subset of $\mathbb{R}.$ Therefore, $f$ induces a Borel measure $\mu_f$ on $\mathbb{R}.$
Lots of unnecessary jargon in your writing.
If $B$ is Borel set in $\mathbb R$ the $f^{-1}(B) \in \mathcal M$ so $\mu_f$ is well defined. All you have to do is to verify that $\mu_f(\emptyset)=0$ and $\mu_f(\bigcup B_n)=\sum \mu_f(B_n)$ for any disjoint sequence of Borel sets $(B_n)$. For this just observe that $(f^{-1}(B_n))$ is a disjoint sequence in $\mathcal M$ and use the fact that $\mu$ is a measure.