Homepage Solution manuals Terence Tao An Introduction to Measure Theory Exercise 1.4.47 (Defect version of Fatou's lemma)

Exercise 1.4.47 (Defect version of Fatou's lemma)

Let (X,,μ) be a measure space, and let (fn)n be a sequence of unsigned absolutely integrable functions that converges pointwise to an absolutely integrable limit f. Show that as n we have

X (fn f )dμ f fnL1 0.

Answers


We have to demonstrate that

lim n (fn f ) = lim n |f fn|

  • From the pointwise dominance fn f |f fn| using the monotonicity of the integral we obtain

    (fn f ) |f fn|

    for each n . Taking the limits preserves the inequality.

  • The first key observation we must make here is that the absolute value function |f fn| can be pointwise repsented using min ,max functions: |f fn| = max (f,fn) min (f,fn) since both f and fn are non-negative. Both max and min are measurable and absolutely integrable; thus, the theorem assertion reduces to

    lim nfn Xfdμ lim nmax (f,fn) + min (f,fn)

    The second observation is that min (f,fn)n f is a sequence of measurable functions dominated by the absolutely integrable function f; thus, by Dominated Convergence Theorem, lim nmin (f,fn) = lim nmin (f,fn) = Xfdμ. Cancelling this term out, we further reduce the theorem assertion to

    lim nfn lim nmax (f,fn)

    Unfortunately we cannot apply the same trick for max (f,fn) since it is not necessarily dominated. We can, however, apply Fatou’s lemma (Corollary 1.4.46) to get

    lim nfnlim nmax (f,fn) [lim nfn lim nmax (f,fn)] = f f = 0

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2020-09-25 00:00
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