1.5. Modes of convergence 101 Anμ(En) → Aμ(E) as n → ∞, which is a consequence of the monotone convergence theorem μ(En) → μ(E) for sets. 1.5.3. Finite measure spaces. The situation simplifies somewhat if the space X has finite measure (and in particular, in the case when (X, B,μ) is a probability space, see Section 2.3). This shuts down two of the four examples (namely, escape to horizontal infinity or width infinity) and creates a few more equivalences. Indeed, from Egorov’s theorem (Exercise 1.4.31), we now have Theorem 1.5.9 (Egorov’s theorem, again). Let X have finite measure, and let fn : X → C and f : X → C be measurable functions. Then fn converges to f pointwise almost everywhere if and only if fn converges to f almost uniformly. Note that when one specialises to step functions using Exercise 1.5.3, then Egorov’s theorem collapses to the downward monotone convergence property for sets (Exercise 1.4.23(iii)). Another nice feature of the finite measure case is that L∞ convergence implies L1 convergence: Exercise 1.5.4. Let X have finite measure, and let fn : X → C and f : X → C be measurable functions. Show that if fn converges to f in L∞ norm, then fn also converges to f in L1 norm. 1.5.4. Fast convergence. The typewriter example shows that L1 conver- gence is not strong enough to force almost uniform or pointwise almost everywhere convergence. However, this can be rectified if one assumes that the L1 convergence is suﬃciently fast: Exercise 1.5.5 (Fast L1 convergence). Suppose that fn,f : X → C are measurable functions such that ∑∞ n=1 fn − f L1(μ) ∞ thus, not only do the quantities fn − f L1(μ) go to zero (which would mean L1 convergence), but they converge in an absolutely summable fashion. (i) Show that fn converges pointwise almost everywhere to f. (ii) Show that fn converges almost uniformly to f. (Hint: If you have trouble getting started, try working first in the special case in which fn = An1En are step functions and f = 0 and use Exercise 1.5.3 in order to gain some intuition. The second part of the exercise implies the first, but the first is a little easier to prove and may thus serve as a useful warmup. The ε/2n trick may come in handy for the second part.) As a corollary, we see that L1 convergence implies almost uniform or pointwise almost everywhere convergence if we are allowed to pass to a subsequence:

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