Multivariable Advanced Calculus

8.7. THE LEBESGUE INTEGRAL, L 1 189and for m, n large enough, this last is given to be small so {I (s n )} is a Cauchy sequencein C and so it converges. This verifies the limit in 8.6 at least exists. It remains toconsider another sequence {t n } having the same properties as {s n } and verifying I (f)determined by this other sequence is the same. By Lemma 8.7.3 and Fatou’s lemma,Theorem 8.5.1 on Page 186,∫|I (s n ) − I (t n )| ≤ I (|s n − t n |) = |s n − t n | dµ∫≤ |s n − f| + |f − t n | dµ∫∫≤lim infk→∞|s n − s k | dµ + lim infk→∞|t n − t k | dµ < εwhenever n is large enough. Since ε is arbitrary, this shows the limit from using the t nis the same as the limit from using s n .Why is L 1 (Ω) a vector space? Let f, g be in L 1 (Ω) and let a, b ∈ C. Then let {s n }and {t n } be sequences of complex simple functions associated with f and g respectivelyas described in Definition 8.7.4. Consider {as n + bt n } , another sequence of complexsimple functions.Theorem 8.6.1,∫∫|as n + bt n − (as m + bt m )| dµ ≤ |a|Then as n (ω) + bt n (ω) → af (ω) + bg (ω) for each ω. Also, from∫|s n − s m | dµ + |b||t n − t m | dµand the sum of the two terms on the right converge to zero as m, n → ∞. Thus af +bg ∈L 1 (Ω). Now here is another characterization for a function to be in L 1 (Ω).Corollary 8.7.6 Let (Ω, S, µ) be a measure space and let f : Ω → C. Then f ∈L 1 (Ω) if and only if f is measurable and ∫ |f| dµ < ∞.Proof: First suppose f ∈ L 1 . Then there exists a sequence {s n } of the sort describedabove attached to f. It follows that f is measurable because it is the limit ofthese measurable functions. Also for the same reasoning |f| = lim n→∞ |s n | so |f| ismeasurable as a real valued function. Now from I being linear,∫ ∫∣ |s n | dµ − |s m | dµ∣ =|I (|s n |) − I (|s m |)| = |I (|s n | − |s m |)| ≤ I (||s n | − |s m ||)∫∫= ||s n | − |s m || dµ ≤ |s n − s m | dµwhich is small whenever n, m are large. As to ∫ |f| dµ being finite, this follows fromFatou’s lemma. ∫∫|f| dµ ≤ lim inf |s n | dµ < ∞n→∞Next suppose f is measurable and absolutely integrable. First suppose f ≥ 0. Thenby the approximation theorem involving simple functions, Theorem 7.7.12, there existsa sequence of nonnegative simple functions s n which increases pointwise to f. Each ofthese must be nonzero only on a set of finite measure because ∫ fdµ < ∞. Note that∫∫∫2f − (f − s n ) dµ + f − s n dµ = 2f