advanced30 min studyLesson 185 in Mathematics

Monotone convergence and Fatou’s lemma

Use order to pass through infinite limits and diagnose why mass can disappear from a pointwise limit.

On this page

What you’ll understand

  • Apply monotone convergence without assuming finite integrals.
  • Derive Fatou’s inequality using increasing tail infima.
  • Construct strict Fatou examples by concentration or escaping support.

An infinite process should not be passed through an integral merely because every finite stage is easy. Two fundamental theorems give controlled permissions. Monotone convergence handles increasing nonnegative functions and even permits an infinite answer. Fatou’s lemma handles arbitrary nonnegative sequences but guarantees only an inequality. Learning why the second conclusion is weaker helps identify the missing hypothesis in many calculations: pointwise limits can lose mass that concentrates into smaller regions or moves away from every fixed observation point.

Increasing nonnegative approximations preserve the integral

Suppose measurable functions fₙ are nonnegative and increase pointwise to f. Monotone convergence states that their integrals increase to the integral of f, with infinity allowed on both sides. There is no requirement that the space have finite measure or that a fixed integrable function dominate the sequence. The order condition prevents previously accumulated nonnegative mass from being removed. The theorem turns the lower-simple-function definition into a practical rule for constructing integrals by increasingly detailed approximations.

A useful consequence is exchanging a nonnegative series with integration: apply the theorem to its partial sums. This works even when the series or resulting integral diverges. If signs are present, partial sums may decrease and cancel; the theorem no longer applies directly. One must establish absolute integrability, a suitable dominating function, or some other theorem with the right assumptions. Naming the reason for the exchange is part of the calculation, especially when the final expression contains an infinite sum.

0≤fn↑f⟹∫f dμ=lim⁡n→∞∫fn dμ;∫∑j=1∞gj dμ=∑j=1∞∫gj dμ(gj≥0)0\le f_n\uparrow f\quad\Longrightarrow\quad\int f\,d\mu=\lim_{n\to\infty}\int f_n\,d\mu;\qquad\int\sum_{j=1}^{\infty}g_j\,d\mu=\sum_{j=1}^{\infty}\int g_j\,d\mu\quad(g_j\ge0)
All functions are measurable. These are equalities of nonnegative extended values, so a divergent result is allowed and involves no subtraction of infinities.↔ Scroll wide equations horizontally.

Tail infima turn a general sequence into an ordered one

For a nonnegative sequence fₙ, define hₙ as the pointwise infimum of fₖ over all k≥n. Dropping early competitors can only increase an infimum, so hₙ increases to the pointwise lower limit of the sequence. Each hₙ is measurable and lies below every remaining fₖ. Monotone convergence applied to hₙ and monotonicity of integration then prove Fatou’s lemma. This construction shows exactly how an apparently unordered sequence is compared with an increasing family of reliable lower bounds.

Fatou’s direction says the integral of the pointwise lower limit cannot exceed the lower limit of the integrals. It does not assert equality and does not guarantee convergence of the original integrals. Nonnegativity can be replaced by a common integrable lower bound after subtracting that bound, but arbitrary negative parts are dangerous. On an infinite-measure space, subtracting a constant may create a nonintegrable adjustment, so the phrase bounded below must not be confused with bounded below by an integrable function.

hn=inf⁡k≥nfk↑lim inf⁡n→∞fn,∫lim inf⁡n→∞fn dμ≤lim inf⁡n→∞∫fn dμh_n=\inf_{k\ge n}f_k\uparrow\liminf_{n\to\infty}f_n,\qquad\int\liminf_{n\to\infty}f_n\,d\mu\le\liminf_{n\to\infty}\int f_n\,d\mu
Fatou applies to nonnegative measurable fₙ. The tail-infimum construction proves the inequality without assuming any convergence of the original sequence.↔ Scroll wide equations horizontally.

Where the missing mass goes

Consider spikes n times the indicator of (0,1/n) on [0,1]. They tend pointwise to zero, including at zero because the interval excludes it, while each integral remains one. Fatou gives zero at most one, which is true and strict. The functions are not increasing: their heights rise while supports shrink. Nonnegativity alone therefore cannot justify equality. This example also rules out any common integrable dominating function, because such a function would permit a stronger limit theorem and contradict the computed areas.

A different mechanism occurs on the whole real line with indicators of [n,n+1]. Each integral is one and each fixed point eventually sees zero, but now mass escapes toward infinity rather than concentrating. This distinction matters when replacing domination by other conditions. A tail-height condition rules out tall spikes but does not by itself stop bounded mass from drifting across an infinite-measure domain. A limit theorem must control the actual mechanism by which an integral could differ from its pointwise prediction.

Integrating a growing stack of nested indicators

On (0,1), set fn(x)=∑j=1n1(0,2−j)(x)f_n(x)=\sum_{j=1}^{n}\mathbf 1_{(0,2^{-j})}(x) and let f be its pointwise limit. Compute the integral of f without finding a closed formula for it.

  1. Each new summand is nonnegative, so fₙ increases pointwise to f. Every fₙ is measurable as a finite sum of measurable indicators.

  2. Integrate the finite sum: its jth indicator contributes length 2⁻ʲ. Thus the integral of fₙ is the geometric sum 1−2⁻ⁿ.

  3. Apply monotone convergence to obtain integral f equal to the limit of 1−2⁻ⁿ, which is one. The increasingly tall stack near zero is still integrable because its total added mass is summable.

The result

The limiting integral is one. Increasing heights alone do not imply divergence; the sizes of the added regions determine the result.

Explore the model

Interactive lab

Interactive lab · function

An ordered family with a visible bound

Choose x=1 and increase the nonnegative coefficient a in y=a x². Predict that the area grows proportionally as a/3, then inspect the area readout. This finite quadratic family illustrates order and monotonicity; it does not animate the indicator stack or prove an infinite limit exchange.

An ordered family with a visible boundThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forFirst identify an actual pointwise ordering; changing height and support in opposite directions need not be monotone.

Check your understanding

For fₙ=n·1(0,1/n) on [0,1], which statement is correct?

Try it yourself

On the real line with length measure, examine fₙ=1[n,n+1]. Find its pointwise limit, its sequence of integrals, and the reason monotone convergence cannot be invoked.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Fix one input x before sending n to infinity; then compare consecutive functions at several inputs.

Compare with the solution

For every fixed real x, the interval [n,n+1] eventually lies entirely to its right, so the pointwise limit is zero. Each interval has length one, hence every integral is one. The sequence is not pointwise increasing because a point can belong to one support and leave the next. Fatou allows the strict inequality 0<1. The discrepancy is escaped support, not a growing pointwise height, which is why infinite-domain limit arguments may also need tightness or spatial tail control.

Take it with you

  • Increasing nonnegative sequences permit equality even with infinite integral values.
  • Fatou follows from increasing tail infima and generally yields only an inequality.
  • Concentration and escape are distinct ways pointwise convergence can lose integral mass.

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