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:

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

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,

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

$c \cdot \int_0^1 s(x) \, dx = 2 \cdot 1 = 2.$
True.
Proof. Let $P =\{ x_0, \ldots, x_n \}$ be a partition of $[a,b]$ such that $s(x) = s_k$ on the $k$ th 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

$\int_a^b s(x) \, dx = \int_{a+c}^{b+c} s(x-c) \, dx. \qquad \blacksquare$
True.
Proof. Since $s(x) < t(x)$ implies $s(x)^3 < t(x)^3$ , the result follows immediately $. \qquad \blacksquare$