254 CHAPTER 11. ABSTRACT MEASURE AND INTEGRATION

which converges to 0. Since {I (|sn|)} is a Cauchy sequence, it is bounded by a constant Cand also {|sn|} is a sequence of simple functions of the right sort which converges pointwiseto | f | and so by definition, ∫

| f |dµ = I ( f ) = limn→∞

I (|sn|)≤C.

This shows the only if part.The more interesting part is the if part. Suppose then that f is measurable and

∫| f |dµ <

∞. Suppose first that f has values in [0,∞). It is necessary to obtain the sequence of com-plex simple functions. By Theorem 11.3.9, there exists a sequence of nonnegative simplefunctions, {sn} such that sn (ω) ↑ f (ω). Then by the monotone convergence theorem,

limn→∞

∫(2 f − ( f − sn))dµ =

∫2 f dµ

and so

limn→∞

∫( f − sn)dµ = 0.

Letting m be large enough, it follows∫( f − sm)dµ < ε and so if n > m∫

|sm− sn|dµ ≤∫| f − sm|dµ < ε.

Therefore, f ∈ L1 (Ω) because {sn} is a suitable sequence.The general case follows from considering positive and negative parts of real and imag-

inary parts of f . These are each measurable and nonnegative and their integral is finite soeach is in L1 (Ω) by what was just shown. Thus

f = Re f+−Re f−+ i(Im f+− Im f−

)and so f ∈ L1 (Ω). This proves the corollary.

Theorem 11.4.9 (Dominated Convergence theorem) Let fn ∈ L1(Ω) and suppose

f (ω) = limn→∞

fn(ω),

and there exists a measurable function g, with values in [0,∞],3 such that

| fn(ω)| ≤ g(ω) and∫

g(ω)dµ < ∞.

Then f ∈ L1(Ω) and ∫f dµ = lim

n→∞

∫fndµ .

3Note that, since g is allowed to have the value ∞, it is not known that g ∈ L1 (Ω) .