Let $f: [a,b] \rightarrow \mathbb{R}$ and $g: [a,b] \rightarrow \mathbb{R}$ Prove the following.
$f \le M \implies \int^b_a f(x)dx \le M(b-a)$
$f+g \in \mathbb{R[a,b]}$ and $\int^b_a [f(x)+g(x)]dx = \int^b_a f(x)dx+ \int^b_a g(x)dx$
$f \le g \implies \int^b_a f(x)dx \le \int^b_a g(x)dx$
$ fg \in \mathbb{R[a,b]}$
If someone can help me with how to begin the proof, that would be great...
(1) Since $f \leq M$ for all $x\in[a,b]$, then $\int_{a}^{b} f = \inf\{U(f,P)\} \leq U(f,P) \leq M(b-a)$.
(2) The converse is not true, take $g = -f$ and pick your favorite non-Riemann integrable function (Dirchlet function for example) for $f$ and $g$. If you actually mean to ask to prove $\int_{a}^{b} f + g$ is integrable if $f,g \in \mathcal{R}[a,b]$ then consider the inequalities $$U(f + g, P) \leq U(f,P) + U(g,P),$$ and$$L(f+g,P) \geq L(f,P) + L(g,P).$$
(3) This is a lot like (1), $f \leq g$ for all $x \in [a,b]$ implies $$\inf\{U(f,P)\} \leq U(f,P) \leq g \leq U(g,P). $$
The last inequality shows $\inf \{ U(f,P) \} \leq \inf \{ U(g,P) \}.$