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Prove the integral of a step function is invariant under translation

by RoRi

For a step function s(x) on an interval [a,b] prove

    \[ \int_a^b s(x) \, dx = \int_{a+c}^{b+c} s(x-c) \, dx. \]

Proof. Let P = \{ x_0, \ldots, x_n \} be a partition of [a,b] such that s(x) = s_k is constant on the open k-th subinterval of the partition.
Then P = \{ x_0 + c, x_1 + c, \ldots, x_n + c \} is a partition of [a+c, b+c] and s(x-c) = s_k for x_{k-1} + c < x < x_k + c. So,

    \[ \int_a^b s(x) \, dx = \sum_{k=1}^n s_k (x_k - x_{k-1}), \]

and,

    \[ \int_{a+c}^{b+c} s(x-c) \, dx = \sum_{k=1}^n s_k (x_k - x_{k-1}). \]

Thus, \int_a^b s(x) \, dx = \int_{a+c}^{b+c} s(x-c) \, dx. \qquad \blacksquare

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