The integral adds, scales, splits at an interior point and respects order. None of these is obvious from the definition, because the definition is a pair of suprema over all partitions and those do not combine simply. Each is proved by choosing partitions for the pieces, refining to a common one, and applying the criterion.

Measuring two adjacent fields separately and adding gives the same answer as measuring the pair together — provided the fence between them is included in both measurements and contributes nothing. That last clause is the only subtle part, and it is why a single point never matters.

Let $f,g$ be integrable on $[a,b]$ and $k$ a constant. Then $f+g$ and $kf$ are integrable with $\int(f+g)=\int f+\int g$ and $\int kf=k\int f$. If $a<c<b$ then $f$ is integrable on both halves and $\int_a^b f=\int_a^c f+\int_c^b f$. If $f\le g$ throughout then $\int f\le\int g$, and consequently $\big\lvert\int f\big\rvert\le\int\lvert f\rvert$.

• Additivity in the integrand is a statement about sums only. There is no product rule: $\int fg$ is unrelated to $\int f\int g$.
• Order passes to the integral: $f\le g$ throughout gives $$. The strict form is not needed anywhere in this course and is not proved.

Work with partitions containing the splitting point, which costs nothing because adding it is a refinement. Such a partition breaks into a partition of each half, and its upper sum is the sum of the two upper sums — the pieces are simply grouped. Taking infima on both sides turns the identity for sums into one for integrals, and the criterion applied to the two halves shows the function is integrable on each.

Adding the point is a refinement, so nothing is lost by insisting on it.

Such a partition is exactly a partition of each half joined at the point.

The pieces of one are the pieces of the other two, grouped.

Both halves inherit a small gap, since both differences are non-negative.

Taking infima over such partitions on both sides gives the identity.

The integral is additive in the integrand and there is no rule for a product.

• The integral is additive in the integrand and in the interval, and scales by a constant.
• Order passes to the integral, and hence the absolute value bound.
• There is no product rule.
• Changing finitely many values changes nothing.

Great work! You now know how to:

• Apply additivity in the integrand and in the interval
• Use the order property and the absolute value bound
• Estimate an integral from a bound on the function
• Say why finitely many values never matter