Measure and integration
Build integrals that handle limits, probability, and functions defined only almost everywhere.
On this page
The question for today
What you’ll understand
- Interpret measurable sets, null sets, and almost-everywhere equality precisely.
- Use a dominating integrable function to justify exchanging a limit and an integral.
- Distinguish pointwise values from equivalence classes in Lp spaces.
An integral should accommodate continuous curves, random variables, and limits of imperfect approximations. Riemann sums begin by partitioning the input axis. Lebesgue integration instead organizes a measurable function by the sizes of sets on which it takes particular values. The reward is a flexible theory of convergence and a shared language for probability and field theory. Its qualifications matter: a set can be dense and still have measure zero, and almost everywhere never means literally everywhere.
A consistent size assigned to sets
A sigma algebra specifies which sets may be measured; it contains the whole space and is closed under complements and countable unions. A measure assigns these sets nonnegative sizes and adds sizes across countably many disjoint pieces. Lebesgue measure agrees with ordinary interval length. A probability measure instead assigns total mass one. These are the same structural rules with different interpretations, so an expectation is an integral over a probability space rather than a separate mathematical species.
A null set has measure zero. Each singleton in the real line has zero Lebesgue measure, so countable additivity makes the rationals null, even though every open interval contains rational points. Density concerns proximity; measure concerns size. Saying a property holds almost everywhere means its exceptional set is null for the specified measure. A point mass probability distribution can give one point probability one, so a Lebesgue-null set is not automatically null for every other measure.
Integrating simple values and controlling limits
For a nonnegative simple function, multiply each value by the measure of its level set and add. The nonnegative Lebesgue integral is obtained by approximating from below with such simple functions. A signed or complex function is integrable when its absolute value has finite integral. This absolute-integrability requirement prevents ambiguous subtraction of two infinities. It also explains why some oscillatory physics integrals require a regulator or distributional interpretation instead of being ordinary Lebesgue integrals.
The dominated convergence theorem provides a powerful limit rule. Suppose measurable functions converge pointwise almost everywhere and a single integrable function g bounds their absolute values almost everywhere. Their integrals then converge to the integral of the limit. The word single is essential: a bound that changes with n and grows without control proves nothing. Monotone convergence is a different theorem for increasing nonnegative sequences and does not require an integrable dominating function.
Norms ignore null changes, but observables may need more structure
For finite p≥1, the Lp norm measures an integrated power of a function's magnitude. Functions that differ only on a null set have zero Lp distance, so elements of Lp are equivalence classes rather than functions with uniquely specified point values. This is useful for energy and mean-square error, but it means evaluation at one point is not generally a well-defined operation on an L2 equivalence class. Continuity or a stronger function space may restore meaningful point evaluation.
Convergence almost everywhere does not ensure convergence of integrals. A spike can grow taller while narrowing fast enough to disappear at each fixed positive point yet retain fixed area. This is the mechanism behind many dangerous exchanges of limits. The same warning appears in machine learning when rare events carry large losses: looking typical point by point does not control expected loss. An integrability or uniform-integrability argument is needed to rule out escaping mass.
Work it through
A disappearing spike with persistent area
On [0,1] with Lebesgue measure, examine . Can its limit be exchanged with its integral?
For any fixed x>0, eventually 1/n<x, so . At x=0 the open interval also gives zero. The sequence therefore converges pointwise to zero everywhere.
Each integral is height n times width 1/n, hence exactly one. The integral of the pointwise limit is zero, so exchanging these operations gives the wrong answer.
If one integrable g dominated every spike, dominated convergence would force the integrals to approach zero. Since they do not, no such integrable dominator exists. The missing hypothesis is visible as concentration near zero.
The L1 distance from zero stays one. Thus pointwise convergence and L1 convergence are genuinely different claims, even on a finite interval.
The integrals remain one while the pointwise limit integrates to zero. Concentration of mass defeats the proposed exchange.
The living notebook
Explore the model
Interactive lab · probability
Concentration and integrated weight
Change success probability p and trial count n in the exact binomial bars. Their discrete masses sum to one while their location and variance change. This illustrates integration with a discrete probability measure; it does not model the continuous spike sequence, approximate Lebesgue integration, or establish a limit theorem.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. 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 forThe measure determines how weights are combined: sum discrete binomial masses, but integrate a continuous density against length.
Pause & reason
Check your understanding
Your turn
Try it yourself
For on [0,1], justify the limit of its integrals without calculating each antiderivative.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Identify the exceptional point and find a fixed integrable bound.
Compare with the solution
The sequence tends to zero at every x<1 and to one at x=1. The singleton {1} has Lebesgue measure zero, so the almost-everywhere limit is zero. Since and the constant one has integral one, dominated convergence gives an integral limit of zero. This agrees with the direct value 1/(n+1).
Keep the idea
Take it with you
- Almost everywhere is a precise measure-dependent qualification, not a synonym for everywhere.
- A fixed integrable dominator makes a particular limit exchange legitimate.
- Lp spaces identify null-set changes and require care with pointwise observables.
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