If $a \lt b$ and $f,g \in R[a,b]$ satisfy $f \leq g$ then $\int^b_af(x)dx \leq \int^b_ag(x)dx$ I am asked to prove the comparison property for regulated functions, namely :
If $a \lt b$ and $f,g \in R[a,b]$ satisfy $f \leq g$ then $\int^b_af(x)dx \leq \int^b_ag(x)dx$.
The definition for regulated function is as follows:

Now, my proof is:
Let $\phi_n,\psi_n \in S[a,b]$ be sequences of step functions converging uniformly to $f$ and $g$ respectively and $\phi_n \leq \psi_n \forall x \in [a,b]$. Let $P=\{p_0,...,p_k\}$ be a partition compatible with both $\phi_n$ and $\psi_n$.
Then $\phi_n \leq \psi_n \Rightarrow f \leq g$ (I think $f \leq g \Rightarrow \phi_n \leq \psi_n$ and hence $\phi_n \leq \psi_n \Leftrightarrow f \leq g$, right?)
Furthermore, $\phi_n \leq \psi_n$
$\Leftrightarrow \sum^k_{i=1}\phi_n(p_i-p_{i-1}) \leq \sum^k_{i=1}\psi_n(p-p_{i-1})$
$\Leftrightarrow \text{lim}_{n \rightarrow \infty}\sum^k_{i=1}\phi_n(p_i-p_{i-1}) \leq \text{lim}_{n \rightarrow \infty} \sum^k_{i=1}\psi_n(p-p_{i-1})$
$\Leftrightarrow \int^b_a f(x)dx \leq \int^b_ag(x)dx$
Is my proof correct?
 A: In your proof it is not clear, why you can choose the step functions $\phi_n$ and $\psi_n$, such that $\phi_n \leq \psi_n$...
So here is my suggestions:
Proposition. Let $a < b$, $f, g \in R[a,b]$ satisfy $f \leq g$.
Then:
$$
\int_a^b f(x) \, \mathrm dx \leq \int_a^b g(x) \, \mathrm dx
$$
Proof.
Let $h := g - f$. Since regulated functions form a vector space and $h \geq 0$, we have
$$
h \in R[a,b] \quad \text{and} \quad 0 \leq h \; .
$$
We now show that
$$
\int_a^b h(x) \, \mathrm dx \geq 0 \; .
$$
For every $n \in \mathbb N^\times$ there is a step function $\varphi_n \in S[a,b]$, such that
$$
\Vert \varphi_n - h \Vert_\infty < \frac 1 n \; .
$$
That implies
$$
- \frac 1 n < \varphi_n(x) - h(x) < \frac 1 n \quad \text{for each } x \in [a,b] \; ,
$$
so that
$$
h(x) -\frac 1 n < \varphi_n(x) \quad \text{for each } x \in [a,b] \; .
$$
Since $h \geq 0$, we get
$$
- \frac 1 n \leq h(x) - \frac 1 n < \varphi_n(x)
$$
for each $x \in [a,b]$.
If $P = \{p_0, \ldots, p_k\}$ is a partition compatible to $\varphi_n$ and $\varphi_{n,i}$ the value of $\varphi_n$ on $(p_{i-1}, p_i)$, then
$$
- \frac 1 n (p_i - p_{i-1}) \leq \varphi_{n,i} \cdot (p_i - p_{i-1})
\quad \text{for each } i \in {1,\ldots,k} \; .
$$
So taking the sum over $i = 1,\ldots,k$ gives
$$
-\frac 1 n \underbrace{\sum_{i=1}^k (p_i - p_{i-1})}_{=(b-a)}
\leq \sum_{i=1}^k \varphi_{n,i} \cdot (p_i - p_{i-1})
=\int_a^b \varphi_n(x) \, \mathrm dx
$$
By taking the limit $n \to \infty$, we get
$$
0 \leq
\lim_{n \to \infty} \int_a^b \varphi_n(x) \, \mathrm dx
= \int_a^b h(x) \, \mathrm dx \; .
$$
Finally, by the linearity of the integral,
$$
0 \leq
\int_a^b g(x) - f(x) \, \mathrm dx = \int_a^b g(x) \, \mathrm dx - \int_a^b f(x) \, \mathrm dx \; ,
$$
i.e.
$$
\int_a^b f(x) \, \mathrm dx
\leq
\int_a^b g(x) \, \mathrm dx \; .
$$
A: As pointed out in the comments, there is an issue because $f\le g$ does not imply $\phi_n \le \psi_n$. To correct it, let's introduce an $\epsilon>0$. Since $\phi_n$ and $\psi_n$ converge uniformly to $f$ and $g$, respectively, there exists $N\in \mathbb{N}$ such that $||f-\phi_n|| < \epsilon$ and $||g-\psi_n|| < \epsilon$ for all $n\ge N$ (where the norms here are the sup norm). Hence we have $f-\phi_n > -\epsilon$ and $g-\psi_n < \epsilon$, which can be rearranged to see that $\phi_n \le \psi_n + 2\epsilon$. 
At this point, apply the same argument you used in your answer to obtain $$
\int_a^b f(x) dx \le \int_a^b g(x) dx + \int_a^b 2\epsilon dx = \int_a^b g(x) dx + 2(b-a)\epsilon
$$
Since $\epsilon$ can be arbitrarily small, you can let it go to 0 and get the desired inequality.

In a nutshell: $f\le g$ doesn't necessarily imply $\phi_n\le \psi_n$, but it does imply $\phi_n$ is almost less than or equal to $\psi_n$.
