# 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?

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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.

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But the conditions of the theorem state that the function has a finite-measure support, which your example does not have. Also, it is easy to prove the quoted line for finite measure sets - which is the case here. –  yaakov Mar 30 '11 at 21:42
:-) Thank you for the credit! I don't think I quite deserved that, though. (lol) –  Giuseppe Negro Mar 30 '11 at 21:43
@yaakov: Oh, good point. So the condition is redundant. –  Nate Eldredge Mar 30 '11 at 23:36

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.

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But the counting measure is not regular. –  yaakov Mar 30 '11 at 21:19
Your $\mu$ is not a regular measure. –  Nate Eldredge Mar 30 '11 at 21:22
Ensuring g is compactly supported is done by the regularity and topological conditions. Specifically, we can approximate (because of regularity, and finite measure of A) A by a compact set, and from local compatibility we can find an open set with a compact closure which includes A. –  yaakov Mar 30 '11 at 21:24
Of course you're right, sorry about that. By the way, what do you exacly mean with "regular"? I recall different books adopting slightly different definitions. –  Giuseppe Negro Mar 30 '11 at 21:26
By regular I (or rather, Rudin), mean exactly the conditions which follow from Riestz theorem: The measure of every set is the infimum on measures of open sets containing it. The measure on every set which is open or finite measured is the supremum on compact sets contained by it. –  yaakov Mar 30 '11 at 21:30