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Good night or day or whatever time of the day you're living in. I'm reading the book "The Elements of Integration and Lebesgue Measure" written by Robert G. Bartle. When he gives the proof to the Lebesgue Dominated Convergence Theorem, he sais: "Be redefining the functions $f_n , f$ on a set of measure $0$ we may assume that the convergence takes place on all of $X$ ".
Can someone help me to understand what does he means with this redefinition, please.

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Suppose you have the result when convergence is pointwise on some set. Then it follows that the result also holds when convergence occurs almost everywhere with respect to the entire space $X$. To see this, let $E \subset X$ be the set on which convergence fails, so that $\mu(E) = 0$ and convergence is pointwise on $X-E$. Then

$$\int_{X} f_{n}d\mu = \int_{X-E} f_{n}d\mu + \int_{E}f_{n}d\mu = \int_{X-E}f_nd\mu + 0$$

since $\mu(E)=0$. But, by our supposition, $\int_{X-E}f_{n} d\mu \to \int_{X-E} f d\mu$. Hence, $$\int_{X}f_{n}d\mu \to \int_{X-E}fd\mu = \int_X f d\mu,$$

again, since $\mu(E)=0$.

The point is just that the integral "doesn't care" about measure $0$ sets. Remember that if $f=g$ almost everywhere, then $\int f d\mu = \int g d\mu$. So we may "redefine" (as your author says) $f_{n}$ as $f_{n} \mathbf{1}_{X-E}$.

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  • $\begingroup$ Sorry if I am incorrect, but do you mean "By our supposition, $\int_{X-E} f_n \, d \mu \rightarrow \int_{X-E} \, f d \mu $"? If not, do you mind explaining how you got the result? $\endgroup$ – Cy L Shih Dec 27 '16 at 2:31

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