Step functions: Using $-|\phi| \leq \phi \leq |\phi|$ conclude $|\int^b_a \phi(x)dx| \leq \int^b_a|\phi(x)|dx$ Let $\phi :[a,b] \rightarrow \Bbb R$ be a step function.
Using $-|\phi| \leq \phi \leq |\phi|$ I need to conclude that $|\int^b_a \phi(x)dx| \leq \int^b_a|\phi(x)|dx$
I have no idea how to prove this. I start with :
Let $P$ be a partition $P=\{p_0,...,p_k\}$ on $[a,b]$ such that it is compatible with $\phi$. Let $\phi_i$ be the values that $\phi$ takes on each interval $(p_{i-1}, p_i)$. We know that $-|\phi| $ and $|\phi|$ are step functions, so we can see that $-|\phi_i| \leq \phi_i \leq |\phi_i|$ for all $1 \leq i \leq k$. Then 
$$-|\phi_i| \leq \phi_i \leq |\phi_i|\text{ \\ $ \times (p_i-p_{i-1})$}$$
$$\Rightarrow -|\phi_i|(p_i-p_{i-1}) \leq \phi_i(p_i-p_{i-1}) \leq |\phi|(p_i-p_{i-1}) \text{\\ $\times \sum^k_{i=1}$}$$
$$\Rightarrow \sum^k_{i=1}-|\phi_i|(p_i-p_{i-1}) \leq \sum^k_{i=1}\phi_i(p_i-p_{i-1}) \leq \sum^k_{i=1}|\phi_i|(p_i-p_{i-1})$$
$$\Rightarrow \int^b_a -|\phi|(x)dx \leq \int^b_a\phi(x)dx \leq \int^b_a|\phi|(x)dx$$
And this is where I get stuck. I think I'm not doing the correct thing. If anyone could help me identify what i'm doing wrong, that would be great. Or just hint me how to start my proof.

\begin{align*}
-|\varphi |\leq \varphi \leq |\varphi |&\implies -\int |\varphi |\leq \int \varphi \leq \int|\varphi |\\
&\implies \left|\int \varphi \right|\leq \left|\int |\varphi |\right|=\int|\varphi |.
\end{align*}
