1.3. The Lebesgue integral 65 there exists a continuous function fn : Rd C for which the set {x B(0,n) : |f(x) fn(x)| 1/n} has measure at most 1 2n . You may find Exercise 1.3.25 below to be useful for this.) Use this (and Egorov’s theorem, Theorem 1.3.26) to give an alternate proof of Lusin’s theorem for arbitrary measurable functions. Remark 1.3.29. This is a trivial but important remark: when dealing with unsigned measurable functions such as f : Rd [0, +∞], then Lusin’s the- orem does not apply directly because f could be infinite on a set of positive measure, which is clearly in contradiction with the conclusion of Lusin’s theorem (unless one allows the continuous function to also take values in the extended non-negative reals [0, +∞] with the extended topology). How- ever, if one knows already that f is almost everywhere finite (which is, for instance, the case when f is absolutely integrable), then Lusin’s theorem applies (since one can simply zero out f on the null set where it is infinite, and add that null set to the exceptional set of Lusin’s theorem). Remark 1.3.30. By combining Lusin’s theorem with inner regularity (Ex- ercise 1.2.15) and the Tietze extension theorem (see §1.10 of An epsilon of room, Vol. I ), one can conclude that every measurable function f : Rd C agrees (outside of a set of arbitrarily small measure) with a continuous func- tion g : Rd C. Exercise 1.3.25 (Littlewood-like principles). The following facts are not, strictly speaking, instances of any of Littlewood’s three principles, but are in a similar spirit. (i) (Absolutely integrable functions almost have bounded support) Let f : Rd C be an absolutely integrable function, and let ε 0. Show that there exists a ball B(0,R) outside of which f has an L1 norm of at most ε, or in other words, that Rd\B(0,R) |f(x)| dx ε. (ii) (Measurable functions are almost locally bounded) Let f : Rd C be a measurable function, and let ε 0. Show that there exists a measurable set E Rd of measure at most ε outside of which f is locally bounded, or in other words, that for every R 0 there exists M such that |f(x)| M for all x B(0,R)\E. As with Remark 1.3.29, it is important in the second part of the exercise that f is known to be finite everywhere (or at least almost everywhere) the result would of course fail if f was, say, unsigned but took the value +∞ on a set of positive measure.
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