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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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2 Answers 2

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

proof of Lusin's Theorem: Let f be a measurable real-valued function on [a; b] and let  > 0 be given. For each n, there is a continuous function hn on [a; b] such that mfx : jhn(x) 􀀀 f(x)j  =2n+2g < =2n+2. Let En = fx : jhn(x)􀀀f(x)j  =2n+2g. Then jhn(x)􀀀f(x)j < =2n+2 for x 2 [a; b]nEn. Let E = S En. Then mE < =4 and hhni is a sequence of continuous, thus measurable, functions that converges to f on [a; b] n E. By Egoro 's Theorem, there is a subset A [a; b] n E such that mA < =4 and hn converges uniformly to f on [a; b]n(E[A). Thus f is continuous on [a; b]n(E[A) with m(E[A) < =2. Now there is an open set O such that O  (E [ A) and m(O n (E [ A)) < =2. Then f is continuous on [a; b] n O, which is closed, and mO < . By Q2.40, there is a function ' that is continuous on (􀀀1;1) such that f = ' on [a; b] n O. In particular, ' is continuous on [a; b] and mfx 2 [a; b] : f(x) 6= '(x)g = mO < . If f is de ned on (􀀀1;1), let 0 = min(=2n+3; 1=2). Then for each n, there is a continuous function 'n on [n + 0; n + 1 􀀀 0] such that mfx 2 [n + 0; n + 1 􀀀 0] : f(x) 6= 'n(x)g < =2n+2. Similarly for [􀀀n􀀀1+0;􀀀n􀀀0]. Linearise in each interval [n􀀀0; n+0]. Similarly for intervals [􀀀n􀀀0;􀀀n+0]. Then we have a continuous function ' de ned on (􀀀1;1) with mfx : f(x) 6= '(x)g < 4 P =2n+2 = .

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3  
formatting of your answer is awful –  userNaN Jul 19 '13 at 8:31
    
Please improve the formatting of your answer (see the editing help). –  Mårten W Jul 19 '13 at 8:34
    
Flagged this post since it is unreadable. –  AD. Jul 19 '13 at 11:21
    
Please check your encoding and learn some LaTeX. –  AD. Jul 19 '13 at 11:22

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