Theorem 1 (Lusin’s theorem). Suppose that $X$ and $Y$ are Polish spaces, that $\mu$ is a finite Borel measure on $X$, that $f:X\to Y$ is Borel measurable, and that $\varepsilon>0$. Then there exists a compact subset $K$ of $X$, with
$$
\mu(X\setminus K)<\varepsilon,
$$
such that the restriction of $f$ to $K$ is continuous.

Definition 1. Suppose that $f$ is a function on the Borel subsets of a metric space $X$ taking values in $[0,\infty]$. We say $f$ is locally finite if for each $x \in X$ there exists a neighborhood $N$ of $x$ with $f(N) < \infty$.

Definition 1. A mapping $f$ from the Borel sets of a metrizable space $(X,\tau)$ to $[0,\infty]$ is tight if $f(K)<\infty$ for each compact $K$ in $X$ and
$$
f(A)=\sup\{f(K):K \text{ compact},\ K\subseteq A\},
\quad \text{for each } A\in \mathcal{B}(X).
$$

Definition 1. If $(X,\tau)$ is a topological space, then the Borel $\sigma$-field $\mathcal{B}$ of $X$ is the $\sigma$-field generated by the open sets of $X$ and we say a measure $\mu$ defined on $(X,\mathcal{B})$ is a Borel measure.