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Properties of an alternate definition of the integral of a step function

Consider if we chose the following definition for the definite integral of a step function:

    \[ \int_a^b s(x) \, dx = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}). \]

Which of the following properties (all valid for the actual definition) would remain valid under this new definition:

  1. \displaystyle{\int_a^b s + \int_b^c s = \int_a^c s}.
  2. \displaystyle{\int_a^b (s+t) = \int_a^b s + \int_a^b t}.
  3. \displaystyle{\int_a^b c \cdot s = c \cdot \int_a^b s}.
  4. \displaystyle{\int_{a+c}^{b+c} s(x) \, dx = \int_a^b s(x+c) \, dx}.
  5. If s(x) < t(x) for all x \in [a,b], then \displaystyle{ \int_a^b s < \int_a^b < t}.

  1. True.
    Proof. Let P_1 = \{ x_0, \ldots, x_m \} be a partition of [a,c] and P_2 = \{ y_0, \ldots, y_n \} be a partition of [c,b] such that s(x) is constant on the open subintervals of P_1 and P_2. Then, let P = P_1 \cup P_2 = \{ x_0, \ldots, x_m, x_{m+1}, \ldots, x_{m+n} \}, where x_m = y_0, x_{m+1} = x_0, \ldots, x_{m+n} = y_n, so P is a partition of [a,b] and s(x) is constant on the open subintervals of P. Then,

        \begin{align*}  \int_a^c s(x) \, dx + \int_c^b s(x) \, dx &= \sum_{k=1}^m s_k^3 (x_k - x_{k-1}) + \sum_{k=1}^n s_k^3 (y_k - y_{k-1}) & (\text{``def." of } \int_a^b s)\\  &= \sum_{k=1}^m s_k^3 (x_k - x_{k-1}) + \sum_{k=m+1}^{n+m} s_k^3 (x_k - x_{k-1}) \\  &= \sum_{k=1}^{m+n} s_k^3 (x_k - x_{k-1}) \\  &= \int_a^b s(x) \, dx. \qquad \blacksquare \end{align*}

  2. False.
    Counterexample: Let s(x) = t(x) = 1 for all x \in [0,1]. Then,

        \[ \int_0^1 (s(x) + t(x)) \, dx = 8 \]

    while

        \[ \int_0^1 s(x) \, dx + \int_0^1 t(x) \, dx = 1 + 1 = 2. \]

    Hence, this property does not hold. (More generally, since (s+t)^3 \neq s^3 + t^3.)

  3. False. Again, this is because (cs)^3 \neq c(s^3). A specific counterexample is given.
    Counterexample: Let s(x) = 1 for all x \in [0,1] and let c = 2. Then,

        \[ \int_0^1 c \cdot s(x) \, dx = (2)^3 = 8, \]

    while,

        \[ c \cdot \int_0^1 s(x) \, dx = 2 \cdot 1 = 2. \]

  4. True.
    Proof. Let P =\{ x_0, \ldots, x_n \} be a partition of [a,b] such that s(x) = s_k on the kth open subinterval of P. 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 on x_{k-1} + c < x < x_k + c. So,

        \[ \int_a^b s(x) \, dx = \sum_{k=1}^n s_k^3 (x_k - x_{k-1}) \qquad \text{and} \qquad \int_{a+c}^{b+c} s(x-c) \, dx = \sum_{k=1}^n s_k^3 (x_k - x_{k-1}). \]

    Thus, indeed we have

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

  5. True.
    Proof. Since s(x) < t(x) implies s(x)^3 < t(x)^3, the result follows immediately. \qquad \blacksquare

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