Simple functions and the integral
Build the Lebesgue integral from measurable level sets and distinguish legitimate cancellation from infinity minus infinity.
On this page
The question for today
What you’ll understand
- Integrate a simple function without double-counting overlapping sets.
- Approximate a nonnegative measurable function from below with simple functions.
- State the absolute-integrability condition for a finite signed integral.
A step function can be integrated by multiplying each height by the size of the set carrying that height. The Lebesgue construction keeps this calculation and broadens what counts as a step: the level sets may be arbitrary measurable sets rather than intervals. Nonnegative functions are then assembled through increasing lower approximations. This order-first approach makes the integral robust under limits and explains why cancellation must be handled only after the positive and negative contributions have been controlled.
A level-set calculation must be representation independent
A nonnegative simple function takes only finitely many nonnegative values on measurable sets. In its canonical representation, those sets are disjoint level sets, so its integral is the sum of value times measure. The convention zero times infinity equals zero is used for a zero level on an infinite-measure set. Refining two descriptions to a common partition proves that the answer does not depend on the representation. The function itself, rather than a particular written sum of indicators, is what gets integrated.
An expression such as two times the indicator of A plus three times the indicator of B need not use disjoint sets. On the overlap the actual height is five. One may refine into A outside B, B outside A, and the intersection, or use the already established linearity of simple-function integration. Counting the overlap only once with height two or three would be wrong. The ability to split measurable regions without changing the answer is the concrete role played by measure additivity.
Approximate values from below
For a nonnegative measurable f, define its integral as the supremum of the integrals of nonnegative simple functions lying below it. The result may be infinite. A useful explicit approximation first truncates f at height n, then rounds downward to the nearest multiple of two to the minus n. This produces finitely many levels and measurable level sets. Because both the cutoff and resolution improve, these simple functions increase pointwise to f, including at points where f is infinite.
The approximation organizes the range into value bands, rather than requiring f to behave smoothly across intervals in the domain. That distinction handles functions such as the indicator of the rationals, which has no ordinary Riemann integral on [0,1] but has Lebesgue integral zero. The definition also yields monotonicity immediately: every lower simple approximation to f is eligible for a larger function g when f≤g. Proving the full limit and linearity theorems requires further arguments, which the next lessons develop.
Signed functions need control before subtraction
For a real measurable function, split it into the positive part and negative part. Both are nonnegative, and their sum is its absolute value. The finite Lebesgue integral is defined when that absolute value has finite integral; then both part integrals are finite and can safely be subtracted. One can also define an extended integral when exactly one part is infinite. If both are infinite, the expression infinity minus infinity is undefined. Oscillation or symmetry does not repair that failure within ordinary Lebesgue integration.
Null modifications leave integrals unchanged: if integrable measurable functions agree almost everywhere, their positive and negative parts have the same respective integrals. On an incomplete measure space, however, changing values on an arbitrary subset of a null set may destroy measurability, so that qualification is not decorative. Once functions are integrable, linearity and the triangle inequality follow, and finite integrals behave like the familiar area calculations. The construction explains both their strength and the exact assumptions behind their algebra.
Work it through
Refining two overlapping indicators
On [0,1] with length measure, take , , and . Compute its integral.
Ignoring finitely many endpoints, the four consecutive quarter-intervals have heights 2, 5, 3, and 0. The overlap contributes the combined height five.
Multiply each height by 1/4 and sum: 2/4+5/4+3/4=10/4=5/2.
As a cross-check, simple-function linearity gives 2m(A)+3m(B)=2·(1/2)+3·(1/2)=5/2. The two descriptions agree because they describe exactly the same function.
The integral is 5/2. Overlapping indicator expressions are valid, provided their summed height is respected on the intersection.
The living notebook
Explore the model
Interactive lab · function
A continuous function behind simple lower bands
Set a=1 and x=1 and predict the quadratic area 1/3, then inspect the readout. Imagine rounding the heights downward into finer value bands to approach that area from below. Vary a while keeping it nonnegative to check scaling. The lab plots the analytic quadratic, not the simple approximations themselves.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
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 forSimple lower approximations explain an integral without requiring the plotted function to be piecewise constant.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let . Find its Lebesgue integral and decide whether it is Riemann integrable.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
For Lebesgue integration use the measures of the two level sets; for Riemann integration inspect every nontrivial subinterval.
Compare with the solution
The rational level set is countable and null, while its complement in [0,1] has measure one. The Lebesgue integral is therefore 1·0+2·1=2. Every subinterval contains both rational and irrational points, so each lower Darboux sum is one and each upper Darboux sum is two. Their gap never shrinks, so the Riemann integral does not exist. The different conclusions follow from different definitions, not from assigning different lengths to intervals.
Keep the idea
Take it with you
- The simple-function integral depends on actual level values and measurable set sizes.
- Nonnegative integration is a supremum of finite-valued lower approximations.
- Finite signed integration requires absolute integrability before cancellation.
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