Pass an increasing limit through the integral

Try

Fill a reservoir one nonnegative layer at a time. If each new layer adds material and never removes earlier material, the final total should equal the limit of the partial totals. Translate that statement into functions before trusting it: f_n(x) must increase at every x, and the functions must be measurable. The measure space itself need not have finite total mass.

Predict

Predict what happens for indicators of [0, n] on the positive real axis. Both the functions and their integrals increase, with an infinite final integral. Now consider indicators of [n, infinity). They decrease instead. Their pointwise limit is zero while every integral is infinite. This contrast identifies the direction of monotonicity as a hypothesis, not decorative wording.

Experiment

On [0,1] use s_n(x)=2^-n floor (2^n x), with any endpoint value at x=1 handled separately. Compute the integrals from the equal-width cells.

n	Number of cells	Integral
1	2	1/4
2	4	3/8
3	8	7/16

Draw the increasing lower staircases beneath y=x. The formula is (1−2^-n)/2, approaching 1/2. Explain why a point at the right endpoint changes no length integral.

Understand

The monotone convergence theorem states: if measurable 0≤f_n↑f, then integral f_n increases to integral f, with infinity allowed. Monotonicity already gives a limit L≤integral f. The work is proving the reverse bound without assuming that the integral is continuous.

Fix a nonnegative finite-valued simple function φ≤f and a number c with 0< c<1. Set E_n to be the measurable set where f_n≥cφ. These sets increase. At any point with φ>0, the limit f is at least φ and therefore strictly exceeds cφ, so the point eventually belongs to E_n. On zero levels the integral contribution is zero regardless.

Consequently integral f_n≥c integral (φ1_(E_n)). Express φ on its finitely many disjoint positive level sets. Continuity below of measure shows that the right-hand integrals increase to integral φ, even when a level set has infinite measure. Thus L≥c integral φ. If any such integral is infinite, L is infinite. Otherwise let c approach one and take the supremum over all φ. The definition of integral f now gives L≥integral f. This completes the proof.

For a nonnegative series Σg_k, partial sums increase to the pointwise extended sum. Finite additivity followed by this theorem yields integral Σg_k=Σ integral g_k. In particular a finite sum of the integrals implies the series is finite almost everywhere: an infinite value on a positive-measure set would force an infinite integral.

Apply

Use this theorem to justify nonnegative additivity for general functions by increasing simple approximations. Name exactly where finite additivity is already known and where a limit is being passed. For decreasing functions, an integrable first function repairs the earlier failure: apply the theorem to f_1−f_n, then subtract only finite quantities. This repair is an early example of controlling mass rather than merely inspecting pointwise behavior.

Continue

An increasing approximation provides both a construction and a proof certificate. Arbitrary oscillating approximations do not. Next, replace monotonicity by a lower-bound argument using tail infima; this produces Fatou’s inequality and reveals how mass may disappear in a pointwise limit. Keep the finite-valued simple minorant step in your own proof outline.

Continue to Bound integrals of lower limits.