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In the third edition of Rudin's *Real and Complex Analysis, Rudin states Lusin's Theorem in an unusual way, and I think there may be an error. Here, $X$ is a locally compact Hausdorff space and $\mu$ is a measure on $X$. Here is his statement:

Suppose $f$ is a complex integrable function on $X$, $\mu(A)<\infty$, $f(x)=0$ if $x\notin A$, and that $\epsilon>0$. Then there exists a $g\in C_{c}(X)$ such that $$ \mu(\{x:f(x)\neq g(x)\})<\epsilon. $$

There is a bit more about some additional conditions we may place on $g$, but the part I am concerned about is in the above statement. To see my problem, consider the case of when $X=\mathbb{R}^{1}$ and $\mu$ is the Lebesgue measure. Let $K$ be a fat Cantor set of positive measure. Then $K$ is compact and totally disconnected. In particular, $K^{c}$ is dense in $\mathbb{R}^{1}$. I think setting $\epsilon$ to be anything less than $m(K)$ gives us a problem here if we are trying to approximate $\chi_{K}$.

On a similar note, any lower semi-continuous function that is bounded above by $\chi_{K}$ will also be bounded above by the zero function. Hence, if $v$ is such a function, we will always have $$ \int_{\mathbb{R}}(\chi_{K}-v)\,dm\geq m(K). $$ However, in his statement of the Vitali-Caratheodory theorem, he states that any $L^{1}$ function can be approximated from below arbitrarily close by lower semi-continuous functions. The function $\chi_{K}$ is obviously in $L^{1}$, so what is going wrong here?

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What problem do you see if $\epsilon<m(K)$? You can approximate $\chi_K$ with piecewise linear continuous functions $f_\epsilon$ such that $\mu\big(\{x:f_\epsilon(x)\ne\chi_K(x)\}\big)<\epsilon$. – Brian M. Scott Mar 14 '12 at 23:28
up vote 8 down vote accepted

Rudin's statement of Lusin's theorem is correct. I suppose you're thinking of $f = I_K$, the indicator function of $K$.
Let $d(x) = \text{dist}(x,K)$ be the distance from $x$ to $K$, which is a continuous function that is 0 iff $x \in K$, and consider the functions $g_n(x) = \max(1 - n d(x),0)$. Then it is not hard to show that $\mu(\{x: f(x) \ne g_n(x)\}) = \mu(\{x: 0 < d(x) < 1/n\}) \to 0$ as $n \to \infty$.

The Vitali-Caratheodory theorem should state that an $L^1$ function can be approximated from below by upper semicontinuous functions and from above by lower semicontinuous functions. Are you sure you're quoting Rudin correctly?

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Oh wow, I got my upper and lowe semicontinuous functions backwards in my counterexample from they way they are in the theorem. Thank you for explaining why $I_{K}$ fails, I appreciate your response. – Alex Lapanowski Mar 14 '12 at 23:23

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