Exercise 47Calculus › Single-variable calculus › Integration › Riemann integralHigh schoolAdded 5 Nov 2026

Linearity of the integral on step functions

#linearity#step functions#sum

Statement

Let $f,g:[a,b]\to\mathbb{R}$ be step functions and $\lambda\in\mathbb{R}$ . Show that $f+\lambda g$ is a step function and that
$\int_a^b(f+\lambda g)=\int_a^bf+\lambda\int_a^bg.$

Hint

Take the common refinement of the two subdivisions, then use linearity of finite sums.

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Solution

Common refinement. Let $\sigma_f$ and $\sigma_g$ be subdivisions adapted to $f$ and $g$ . Take $\sigma=\sigma_f\cup\sigma_g$ (the union of all subdivision points). On each interval of $\sigma$ , both $f$ and $g$ are constant, say $f=u_k$ , $g=v_k$ .

Compute.
$\int_a^b(f+\lambda g)=\sum_k(u_k+\lambda v_k)(x_k-x_{k-1})=\sum_ku_k\Delta_k+\lambda\sum_kv_k\Delta_k.$
This is $\int f+\lambda\int g$ .

Conclusion. The integral is linear on the space of step functions.

Linearity of the integral on step functions

Statement

Solution

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