Prove some properties of averages of functions

Prove that the average $A(f)$ has the following properties:

Additive property: $A(f+g) = A(f) + A(g)$ .
Homogeneous property: $A(cf) = c \cdot A(f)$ for a constant $c \in \mathbb{R}$ .
Monotone property: $A(f) \leq A(g)$ if $f \leq g$ for all $x \in [a,b]$ .

Proof. We compute using the formula for the average of a function on an interval,
$\begin{align*} A(f+g) &= \frac{1}{b-a} \int_a^b (f(x) + g(x)) \, dx \\ &= \frac{1}{b-a} \left( \int_a^b f(x) \, dx + \int_a^b g(x) \, dx & (\text{Additive prop of integrals})\\ &= \frac{1}{b-a} \int_a^b f(x) \, dx + \frac{1}{b-a} \int_a^b g(x) \, dx \\ &= A(f) + A(g). \qquad \blacksquare \end{align*}$
Proof. Again, we compute,
$\begin{align*} A(cf) &= \frac{1}{b-a} \int_a^b c \cdot f(x) \, dx \\ &= \frac{c}{b-a} \int_a^bf(x) \, dx \\ &= c \left( \frac{1}{b-a} \int_a^b f(x) \, dx \right) \\ &= c \cdot A(f). \qquad \blacksquare \end{align*}$
Proof. Assume $f(x) \leq g(x)$ for all $x \in [a,b]$ . Then, by the monotone property of the integral we have,
$\begin{align*} \int_a^b f(x) \, dx \leq \int_a^b g(x) \, dx && \implies && \frac{1}{b-a} \int_a^b f(x) \, dx &\leq \frac{1}{b-a} \int_a^b g(x) \, dx \\ && \implies && A(f) &\leq A(g). \qquad \blacksquare \end{align*}$

Stumbling Robot