Lusin Theorem conditions

Question

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∖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?

Answer 1

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 dissonance's example, 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.

Answer 2

Yes, that's a necessary condition. I believe that there are deep reasons for that, but I'd think at a quick one: if it wasn't for finiteness on compact sets, how could you ensure that $g$ is compactly supported?

Example. EDIT: Wrong. This example lacks regularity on $\mu$.

Let $X=\mathbb{R}$ and $\mu$ the counting measure. Let $f(x)=\chi_{\{0\}}(x)$. That's a measurable $f$ that vanishes outside a set of finite measure $1$. Can we find an approximating $g$ such as the one in the statement of the theorem? I think we cannot.