Define integration from simple functions

Try

A sensor takes value 2 on a measurable region of area 3 and value 5 on a disjoint region of area 1. Its total weighted area is 11. Change the second region to overlap the first: adding two indicator functions gives value 7 on the overlap, rather than replacing one reading with another. Draw the disjoint cells before calculating. The Lebesgue integral begins with these weighted areas; it does not require the regions to be intervals.

Predict

Predict the integral of a reading that equals infinity at one point and zero elsewhere under length measure. Predict it again under counting measure. These answers differ because an integral depends on both the function and the measure. A null set can carry arbitrarily large values without contributing area, whereas a point has mass one under counting measure. Record which measure you used.

Experiment

Construct a common partition for two overlapping regions with μ(A)=3, μ(B)=2 and μ(A∩B)=1.

Cell	Measure	Value of 2·1_A+5·1_B	Contribution
A outside B	2	2	4
B outside A	1	5	5
Intersection	1	7	7

The integral is 16. Check that 2·3+5·2 gives the same result. Then change only the intersection measure: explain why the total remains fixed when the separate measures remain fixed.

Understand

A nonnegative simple function s has finitely many finite values a_j on disjoint measurable sets E_j. Define its integral as the sum of a_j μ(E_j), using zero for a zero coefficient even if the set has infinite measure. To prove independence from representation, refine two partitions to all intersections. Finite additivity separates each original term into refined cells. On a nonempty intersection both coefficients equal the actual function value, so both sums coincide. No infinity is subtracted.

For a measurable extended nonnegative f, define its integral as the supremum of the integrals of all finite-valued simple s with 0≤s≤f. This may be infinity. The collection is nonempty because it contains zero. If f≤g, every minorant of f is a minorant of g, proving monotonicity. Positive scaling follows by the bijection s↔as for a>0; the zero case is separate.

Changing f on a null set does not change this supremum: delete that set from every simple minorant, which preserves its integral, and compare with the other function. In particular infinity on a null set causes no difficulty. Conversely, integral f=0 forces every set where f≥1/k to be null, hence forces f=0 almost everywhere. This is a statement about nonnegative functions; signed cancellation requires separate treatment.

Simple functions are linear under integration by a common refinement. Extending additivity to arbitrary nonnegative functions will use the monotone convergence theorem next. We have not silently assumed an interchange of limit and integral.

Apply

Apply the definition to the indicator of rational points in [0,1]. Its integral is zero, although every interval meets its support. This shows why partitioning the range into measurable level sets differs from requiring narrow vertical rectangles over interval subdivisions. Explain why the pointwise maximum and sum of two indicators have different integrals on their overlap. For a new nonnegative function, identify useful simple minorants before attempting any limit.

Continue

Keep three distinctions in your notes: measurable versus integrable, zero everywhere versus zero almost everywhere, and finite simple values versus an extended nonnegative function. The next theorem proves that increasing simple approximations recover the defining supremum and licenses infinite nonnegative sums. Complete the practice by naming the measure and any null sets explicitly.

Continue to Pass an increasing limit through the integral.