98 1. Measure theory convergence, one cannot simply rank these modes in a linear order from strongest to weakest. This is ultimately because the different modes react in different ways to the three “escape to infinity” scenarios described above, as well as to the “typewriter” behaviour when a single set is “overwritten” many times. On the other hand, if one imposes some additional assumptions to shut down one or more of these escape to infinity scenarios, such as a finite measure hypothesis μ(X) ∞ or a uniform integrability hypothesis, then one can obtain some additional implications between the different modes. 1.5.1. Uniqueness. Throughout these notes, (X, B,μ) denotes a measure space. We abbreviate “μ-almost everywhere” as “almost everywhere” throughout. Even though the modes of convergence all differ from each other, they are all compatible in the sense that they never disagree about which function f a sequence of functions fn converges to, outside of a set of measure zero. More precisely: Proposition 1.5.7. Let fn : X → C be a sequence of measurable functions, and let f, g : X → C be two additional measurable functions. Suppose that fn converges to f along one of the seven modes of convergence defined above, and fn converges to g along another of the seven modes of convergence (or perhaps the same mode of convergence as for f). Then f and g agree almost everywhere. Note that the conclusion is the best one can hope for in the case of the last five modes of convergence, since as remarked earlier, these modes of convergence are unaffected if one modifies f or g on a set of measure zero. Proof. In view of Exercise 1.5.2, we may assume that fn converges to f either pointwise almost everywhere, or in measure, and similarly that fn converges to g either pointwise almost everywhere, or in measure. Suppose first that fn converges to both f and g pointwise almost ev- erywhere. Then by Exercise 1.5.1, 0 converges to f − g pointwise almost everywhere, which clearly implies that f − g is zero almost everywhere, and the claim follows. A similar argument applies if fn converges to both f and g in measure. By symmetry, the only remaining case that needs to be considered is when fn converges to f pointwise almost everywhere, and fn converges to g in measure. We need to show that f = g almost everywhere. It suﬃces to show that for every ε 0, that |f(x) − g(x)| ≤ ε for almost every x, as the claim then follows by setting ε = 1/m for m = 1, 2, 3,... and using the fact that the countable union of null sets is again a null set.

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