If $f(x)\leq g(x)$ for all $x\in [a,b]$ then $\displaystyle \int_a^b f(x)dx\leq \int_a^b g(x)dx$ I'm having trouble constructing a proof for the following theorem:

Let $f$ and $g$ be Riemann integrable functions on $[a,b]$. Show that:  if $f(x)\leq g(x)$ for all $x\in [a,b]$ then:
$$\int_a^b f(x)dx\leq \int_a^b g(x)dx$$

Any suggestions on how to start, or an answer would be appreciated!
 A: Take the lower Riemann sum with respect to some partition $P$ given by the set of points $x_0<x_1<...<x_{n-1}$, then
$$
L(h,P)=\sum_{i=0}^{n-1}\inf_{t\in [x_{i+1},x_i]}h(t)(x_{i+1}-x_{i})
$$
If $h>0$ on $[a,b]$, what can you tell me about each of the infima above? What about the sign of $x_{i+1}-x_i$? What about the sum of a bunch of positive things? 
A: Let $h(x) = g(x) -f(x)$. Then $h(x) \ge 0$ on $[a,b]$
and we have 
$$\int_a^bh(x)dx = \lim_{n\to \infty}\frac{b-a}{n}\sum_{k=0}^{n-1}h\left(a+(b-a)\frac{k}{n} \right) $$
Note that $$A(n) = \frac{b-a}{n}\sum_{k=0}^{n-1}h\left(a+(b-a)\frac{k}{n} \right) \ge 0$$
since $h(x) \ge 0$ on $[a,b]$.
Then we have, $\lim_{n \to \infty} A(n) \ge 0$.
That is, $$\int_a^bh(x)dx \ge 0 $$
$$\Rightarrow \int_{a}^b(g(x)-f(x))dx \ge 0$$
$$\Rightarrow \int_a^b g(x)dx \ge \int_a^bf(x)dx$$
A: $$f(x)\leq g(x)\Rightarrow f(x)-g(x)\leq 0\; \; on \; [a,b] $$
$$f(x)= f(x)-g(x)+g(x)$$
$$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(x)-g(x)+g(x)dx=\int_{a}^{b}[f(x)-g(x)]dx + \int_{a}^{b}g(x)dx$$
$$\Rightarrow \int_{a}^{b}f(x)dx-\int_{a}^{b}g(x)dx=\int_{a}^{b}[f(x)-g(x)]dx \leq 0$$
$$\Rightarrow \int_{a}^{b}f(x)dx \leq  \int_{a}^{b}g(x)dx$$
A: As Matthew Leingang suggested in the comments, it's sufficient to show that $\int g - f = \int g - \int f \geq 0$. 
Suppose a function $h$ is integrable and $h \geq 0$. Then $\exists$ series of upper and lower step functions, $(\psi)_{n \in \mathbb{N}}$ and $(\phi)_{n \in \mathbb{N}}$, ($\psi_n \leq h \leq \phi_n$) whose integrals converge to $\int h$. In particular, 
$lim_{n \rightarrow \infty} \int \phi_n = \int h$.  
We know that $\phi_n \geq h \geq 0$. Then $\int \phi_n \geq 0$ (since the integral of a step function with partition $\{ x_1, x_2, ..., x_n \}$ is simply $\sum_{i = 1}^{n - 1} (\phi_n(x)_{x \in (x_i, x_{i+1})})(x_{i+1} - x_i)\geq 0$ if $\psi_n \geq 0$). Then it follows that $\int h = lim_{n \rightarrow \infty} \int \psi_n \geq 0$.
Note that if $f$ and $g$ are integrable, then $f-g$ is an integrable function. Replace $h$ in the proof with $f-g$, and you are done.
A: 
Claim: If $h$ is Riemann integrable on $[a,b]$ and $h\geq 0$ on the interval $[a,b]$. Then $\int_a^bh\geq 0$.

Let $P=\{x_0,x_1,\dots,x_n\}$ be a partition of $[a,b]$. This means that $$a=x_0<x_1<\dots<x_n=b.$$  For each $i=1,\dots,n$, we have $h(x)\geq 0$ for all $x\in[x_{i-1},x_i]$ and hence, we get
$$m_i=\inf_{x\in[x_{i-1},x_i]}h(x)\geq 0$$
Thus,
$$L(P,h)=\sum_{i=1}^nm_i(x_i-x_{i-1})\geq 0$$
This implies that $L(P,h)\geq 0$ for all $P\in\mathcal{P}[a,b]$, where $\mathcal{P}[a,b]$ denotes the family of all partitions of $[a,b]$. Hence, 
 $$0\leq \overbrace{\sup_{P\in\mathcal{P}[a,b]}L(P,h)}^{\text{Lower Riemann Integral}}=\int_{a}^bh.$$
This proves the claim.


Going back to your question, set $h(x)=g(x)-f(x)$ for all $x\in[a,b]$. Then $h(x)\geq 0$ for all $x\in[a,b]$. Using properties of the Riemann integrals and the above claim, we get
  $$\int_a^bg-\int_a^bf=\int_a^b(g-f)=\int_a^bh\geq 0.$$
  Done. Hope this helps.

