Monotone Convergence Theorem

Given a sequence of functions {fn}{fn} that converges pointwise to some limit function f, it is not always true that limnfn=limnfn. (Take this sequence for example.) The Monotone Convergence Theorem (MCT), the Dominated Convergence Theorem (DCT), and Fatou's Lemma are three major results in the theory of Lebesgue integration that answer the question, "When do limn and commute?" The MCT and DCT tell us that if you place certain restrictions on both the fn and f, then you can go ahead and interchange the limit and integral. Fatou's Lemma, on the other hand, says, "Here's the best you can do if you don't make any extra assumptions about the functions."

Last week we discussed Fatou's Lemma. Today we'll look at an example that uses the MCT. And next week we'll cover the DCT.

Monotone Convergence Theorem: If {fn:X[0,)} is a sequence of measurable functions on a measurable set X such that fnf pointwise almost everywhere and f1f2, then limnXfn=Xf.  

Let's look at an example that, on the surface, looks quite nasty. But thanks to the MCT, it's not bad at all.

Example

Let X be a measure space with a positive measure μ and let f:X[0,] be a measurable function. Prove that limnXnlog(1+fn)dμ=Xfdμ.

Proof. Begin by defining
fn=nlog(1+fn)=log(1+fn)n
and note that each fn is nonnegative (since both log and f are nonnegative) and measurable (since the composition of a continuous function with a measurable function is measurable). Further, f1f2. Indeed, log is an increasing function and for a fixed xX the sequence (1+f(x)n)n is increasing. In fact*, it increases to ef(x). In other words, limnfn(x)=limnlog(1+fn)n=logef(x)=f(x).
Hence, by the Monotone Convergence Theorem
limnXfndμ=xfdμ
as desired.

Not so bad, huh?

Remark

Did you know that the MCT has a "continuous cousin"? (Well, maybe it's more like a second cousin.) Have you come across Dini's Theorem before?

Dini's Theorem: If {fn:XR} is a nondecreasing sequence of continuous functions on a compact metric space X such that fnf pointwise to a continuous function f:XR, then the convergence is uniform. 

Here we have a monotone sequence of continuous—instead of measurable—functions that converge pointwise to a limit function f on a compact metric space. By Dini's Theorem, the convergence is actually uniform. So IF the fn are also Riemann integrable, then we can conclude** limnXfn=Xf.

Perhaps this doesn't surprise us too much: we've seen before that continuity and measurability are analogous notions (to a certain extent)!   


Footnotes

*Recall from elementary calculus: limn(1+xn)n=ex for any xR

** See Rudin's Principles of Mathematical Analysis (3ed.), Theorem 7.16.

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