The conditions on Lusin's Theorem and its Corollary

Question

Here is the Lusin's Theorem

Let $X$ be a locally compact Hausdorff space and let $μ$ be a Riesz measure on $X$. Suppose $f$ is a measurable function on $X$, $μ(A)<∞$, $f(x)=0$ if $x∈X \setminus A$, and $ϵ>0$. Then there exists a continuous function $g$ on $X$ with compact support such that $μ(x:f(x)≠g(x))<ϵ$.

Is the theorem fail when the support of the measurable function $f$ is of infinite measure? Do we have a example?

And here is its Corollary

Assume we have same setting as above and $|f| \leq 1$, then there is a sequence of function $g_i$ in $C_c(X)$ with $|g_i| \leq 1$ such that $\lim_{i\rightarrow \infty}g_i=f_i$ a.e.

It is required that $f$ is bounded, if $f$ is unbounded, is this corollary fail?

Answer 1

If you remove $\mu (A)<\infty $, then "compact support" may fail. As in $X=\mathbb R $, $f=1$.

If $X $ is a countable union of compact ($\sigma $-compact), then one applies the compact version with $\varepsilon/2^{n} $ and the result holds without finite support for $f $ nor compact support for $g $.

But if $X$ is not $\sigma $-compact, it looks like there should be a counterexample; I cannot think of one right now, though.

As for the corollary, the point is that if $f $ is bounded, so are the $g_i $.