Exercise 50Calculus › Single-variable calculus › Integration › Riemann integralHigh schoolAdded 5 Nov 2026

Two subdivisions for the same function

#step function#subdivision#independence

Statement

Let $f:[0,2]\to\mathbb{R}$ be the step function associated with the subdivision $\sigma=\{0,1,2\}$ with values $c_1=2$ , $c_2=5$ on the two subintervals. Show that the "integral" $\sum_k c_k(x_k-x_{k-1})$ does not depend on the subdivision.

Hint

Refine the subdivision: replace $[0,1]$ by $[0,1/2]\cup[1/2,1]$ , both with value $2$ . Compare.

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Solution

Original subdivision. $\int f=2\cdot1+5\cdot1=7$ .

Refined subdivision. $\sigma'=\{0,1/2,1,2\}$ with values $(2,2,5)$ :
$2\cdot\tfrac12+2\cdot\tfrac12+5\cdot1=1+1+5=7.$

General principle. If $\sigma'\supseteq\sigma$ refines $\sigma$ , the subinterval $[x_{k-1},x_k]$ is split into $[x_{k-1},y]\cup[y,x_k]$ with the same value $c_k$ , contributing $c_k(y-x_{k-1})+c_k(x_k-y)=c_k(x_k-x_{k-1})$ : the sum is unchanged.

Conclusion. The integral of a step function is well-defined.

Two subdivisions for the same function

Statement

Solution

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