core29 min studyLesson 177 in Mathematics

Exchanging limits and integrals

Match the mode of convergence to the measure, interval, and tail control required by integration.

On this page

What you’ll understand

  • Derive the uniform-error integral bound on a finite interval.
  • Detect concentration and escaping tails that invalidate a limit exchange.
  • State dominated convergence as a separate measure-theoretic criterion.

An integral accumulates contributions across a domain, so small error at each fixed location may not control total error. A narrowing spike can hide a fixed amount of area, and a broad low plateau can carry its area toward infinity. These are distinct ways an apparently harmless limit can defeat integration. The right theorem must match both the convergence mode and the domain: a finite-interval uniform bound is one sufficient route, while an integrable dominating function is another.

Finite length turns uniform error into area error

Let Riemann-integrable functions converge uniformly on a fixed finite interval. Their uniform limit is also Riemann integrable, and the integral of the difference is bounded by interval length times its supremum magnitude. This yields convergence of the integrals without locating any maximum error or calculating antiderivatives. The bound is quantitative: a tolerance on the desired integral determines a stricter tolerance for the function approximation. The interval must be the same for all functions.

The factor of interval length explains why the same argument cannot be carried over unchanged to an infinite interval. A uniform error of one over n across an interval of length n can retain unit area. Even on a finite interval, pointwise convergence alone gives no common height bound shrinking to zero. Before exchanging symbols, ask whether the integral error is controlled in a norm that actually tends to zero, rather than merely assuming that pointwise agreement must accumulate harmlessly.

∣∫abfn−∫abf∣≤∫ab∣fn−f∣≤(b−a)∥fn−f∥∞⟶0\left|\int_a^b f_n-\int_a^b f\right|\le\int_a^b|f_n-f|\le(b-a)\|f_n-f\|_\infty\longrightarrow0
Assume a<b are finite and fₙ are Riemann integrable on the common interval, with uniform convergence to f.↔ Scroll wide equations horizontally.

Concentration and escape are two different warnings

On the unit interval, give a function height n on the open interval from zero to one over n and zero elsewhere. Every fixed positive input eventually lies outside that support, and the value at zero is always zero. Thus the functions tend pointwise to zero everywhere, but their integrals remain one. Mass concentrates in a narrowing region rather than disappearing. Each function is Riemann integrable, so the failure concerns the interchange, not the existence of the individual integrals.

On the nonnegative real line, instead use height one over n on the interval from zero to n. These functions tend uniformly to zero because their supremum heights vanish, yet their integrals again remain one. Here the error spreads over a domain whose effective length grows without bound. Finite-domain uniform convergence prevents neither scenario automatically when the hypotheses are changed: the spike violates uniform convergence, while the plateau violates the finite-length assumption of the earlier estimate.

fn=n1(0,1/n):fn→0 pointwise on [0,1],∫01fn=1f_n=n\mathbf1_{(0,1/n)}:\quad f_n\to0\ \text{pointwise on }[0,1],\quad\int_0^1 f_n=1
The shrinking support has length 1/n, exactly compensating for its height n. Assigning zero at the endpoints makes pointwise convergence hold everywhere.↔ Scroll wide equations horizontally.

A common integrable envelope gives another route

Lebesgue dominated convergence applies to measurable functions fₙ on one measure space when fₙ tends to f almost everywhere and one integrable function g bounds every |fₙ| almost everywhere. The envelope g may not depend on n. It rules out uncontrolled concentration and escaping tails in a way suited to the measure space, and it yields convergence in integrated absolute error. Almost everywhere refers to the chosen measure; changing that measure can change which exceptions are negligible. This theorem will receive its construction and proof in the later measure-theory module.

For example, xⁿ on the unit interval is bounded by one and tends to zero except at the single endpoint one. Its integrals therefore tend to zero even though convergence is not uniform on that interval. The quadratic lab supports the simpler finite-interval estimate: predict how both its value and its displayed accumulated area change when a shrinks at fixed x. It cannot certify a Lebesgue dominator or reveal arbitrary concentrated spikes; those require an analytic argument outside its small model.

Uniform convergence is sufficient but not necessary

Evaluate the limit of ∫01xn dx\int_0^1x^n\,dx and explain which convergence argument is valid.

  1. For every x<1, xⁿ tends to zero, whereas at x=1 it stays one. The pointwise limit differs from zero only on one Lebesgue-null point.

  2. The constant one is an integrable bound independent of n. Dominated convergence therefore implies that the integrals approach zero, although uniform convergence on [0,1] fails.

  3. A direct calculation gives ∫01xn dx=1/(n+1)\int_0^1x^n\,dx=1/(n+1), confirming the limit and providing an explicit finite-n error. The integral of the pointwise limit is also zero.

The result

The integral limit is zero. Dominated convergence, rather than a nonexistent uniform bound on [0,1], justifies passing the limit through this integral.

Explore the model

Interactive lab

Interactive lab · function

A shrinking coefficient controls the area

Fix an inspected endpoint x and vary a in y=ax². Predict that the integral from zero to x scales linearly with a, alongside the domain error bound |a|R² on a fixed [−R,R]. The lab displays that exact quadratic integral, not a general limit-exchange test or spike model.

A shrinking coefficient controls the areaThis 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 forAn explicit integrable error bound connects pointwise information to accumulated error.

Check your understanding

Which missing finite-interval hypothesis explains why n⁻¹1_[0,n] can converge uniformly to zero but retain integral one on [0,∞)?

Try it yourself

On [0,2], suppose continuous fₙ converge uniformly to f with ∥fn−f∥∞≤3/n2\|f_n-f\|_\infty\le3/n^2. How many terms guarantee integral error at most 0.01?

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

Give me a hint

Multiply the supremum error by the interval length before solving for n.

Compare with the solution

The absolute integral error is at most 2·3/n²=6/n². Requiring 6/n²≤0.01 means n²≥600, so n=25 suffices while n=24 does not meet this bound. This is a guarantee based on the supplied envelope, not a claim that the actual error necessarily reaches it. Uniform convergence also ensures the continuous limit is integrable on this finite interval.

Take it with you

  • Finite interval length converts a uniform error bound into an integral error bound.
  • Pointwise concentration and infinite-domain tail escape are separate failure modes.
  • Dominated convergence uses a single integrable envelope and can apply without uniform convergence.

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