Bound integrals of lower limits

Try

Imagine observing a sequence of nonnegative density maps. Each map has total mass one, but the bright region becomes narrower and taller. If every fixed location eventually becomes dark, does the limiting map still have total mass one? Calculate the example f_n=n·1_(0,1/n), defined as zero at x=0, on the unit interval. Its pointwise limit is zero. Its integral is always one.

Predict

Predict the direction of a universally valid inequality. The limit can lose visible mass, so its integral should be no larger than a lower limiting total. It need not equal that total. Ask what happens if the same spike is made negative. The negative example reverses the obstruction and will explain why a lower control assumption matters.

Experiment

Track the moving positive spike using an exact table; the heights and widths are separate quantities.

n	Height	Width	Integral	Value at x=1/3
2	2	1/2	1	2
4	4	1/4	1	0
8	8	1/8	1	0

Choose another fixed positive x and find an index after which its reading is zero. Then compare with an interval that depends on n. Pointwise convergence fixes x first; allowing the observation location to move is a different statement.

Understand

For a numerical sequence a_n, its lower limit is the increasing limit of infima of its tails. Fatou’s lemma says that measurable nonnegative f_n satisfy integral (liminf f_n)≤liminf integral f_n. Values and integrals may be infinite.

Define g_n(x)=inf over k≥n of f_k(x). A countable infimum of measurable functions is measurable, and deleting early terms can only increase this infimum. Thus g_n increases to liminf f_n. Also g_n≤f_k for every k≥n, so integral g_n≤inf over k≥n of integral f_k. Apply monotone convergence to the left and the definition of lower limit to the right. This proves Fatou without exchanging an arbitrary limit and integral.

The inequality can be strict, as the spike shows. It is a lower semicontinuity statement, not a conservation law. If a sequence converges almost everywhere, modifying all functions on the single union of their exceptional null sets preserves the integrals and allows the same argument. For p>0, applying the lemma to abs (f_n)^p bounds the integral of abs (f)^p by the lower limiting integrals whenever f_n→f almost everywhere.

For signed functions, a common integrable lower bound h permits applying the nonnegative theorem to f_n−h. The integrable term can then be restored without an infinity-minus-infinity expression. Similarly, if f_n≤g for one integrable g and the relevant signed integrals are defined, applying Fatou to g−f_n yields a reverse upper-limit inequality. Both bounds together will be the engine of dominated convergence.

Apply

Replace the positive spike with −n·1_(0,1/n). Its limiting integral is zero but its integrals are −1, disproving an unrestricted signed version. Next replace concentration by escape: indicators of [n, n+1] on the real line also lose their entire mass in the pointwise limit. Explain why restricting the domain to a finite-measure space prevents this particular escape example but does not prevent concentration.

Continue

Use Fatou when a bound on a limit is enough; do not report it as norm convergence. In the next lesson, an integrable dominating function prevents both concentration and escape and makes the upper and lower bounds agree. Your proof should state nonnegativity or the precise integrable lower control before invoking this lemma.

Continue to Control limits with an integrable envelope.