Lusin Theorem conditions Lusin Theorem (as stated by Rudin):
Let $X$ be a locally compact Hausdorff space and let $μ$ be a regular Borel measure on $X$ such that $μ(K)<∞$ for every compact $K⊆X$. Suppose $f$ is a complex measurable function on $X$, $μ(A)<∞$, $f(x)=0$ if $x∈X \setminus A$, and $ϵ>0$. Then there exists a continuous complex function $g$ on $X$ with compact support such that
$μ(x:f(x)≠g(x))<ϵ$.
But I can't seem to find in the proof anywhere a use of the fact that the measure is finite for compact sets.
Is the condition nessecary? Is there a counter-example?
 A: It can be shown that any $\sigma$-finite regular measure on an LCH space must have $\mu(K) < \infty$ for $K$ compact.  (Exercise: Prove it.)  So we will have to use a measure which is not $\sigma$-finite.  Modifying Giuseppe Negro's example (now deleted), let $X = \mathbb{R}$ and $\mu(A) = \infty$ for all nonempty $A$, which I think is a regular measure (rather trivially so), and take $f$ to be any discontinuous function.
Rudin's proof of Lusin contains the following line:

Fix an open set $V$ such that $A \subset V$ and $\bar{V}$ is compact.  There are compact sets $K_n$ and open sets $V_n$ such that $K_n \subset T_n \subset V_n \subset V$ and $\mu(V_n - K_n) < 2^{-n} \epsilon$.

This invokes Theorem 2.17 (a) (paraphrased):

For any measurable set $E$ and $\epsilon > 0$, there exists $F$ closed and $V$ open with $F \subset E \subset V$ and $\mu(V-F) < \epsilon$.

The proof of 2.17 uses the assumption that $\mu$ is finite on compact sets in the second line, when it asserts that $\mu(K_n \cap E) < \infty$.
Also, it's easy to see that all of Rudin's quoted claims fail using the example I gave.
