Show $\int_E {(f_1 + f_2)d\mu } = \int_E {f_1 d\mu } + \int_E {f_2 d\mu } $ In my textbook, given a measure space $(\Omega,F,\mu)$, the integration for a non-negative $F$ measurable function $f$ on $E$ is defined as $$\int_E f\ \mathsf d\mu  = \sup_{0 \le h \le f} I_E\left( h \right)$$ where $h$ is simple function on $E$, i.e. $$h = \sum_{k = 1}^N {a_k}{\chi _{{E_k}}}\left( x \right)$$ and $${I_E}\left( h \right) = \sum_{k = 1}^N {a_k}\mu \left( {{E_k}} \right).$$
Now the question is to show, using a previous result $I_E(g+h)=I_E(g)+I_E(h)$:

$$\int_E {(f_1 + f_1)\ \mathsf d\mu }  = \int_E {f_1\ \mathsf d\mu }  + \int_E {f_2\ \mathsf d\mu } $$
  where $f_1$ and $f_2$ are all non-negative $F$-measurable functions.

I am able to show $$\int_E \left( {{f_1} + {f_2}} \right)\ \mathsf d\mu  \ge \int_E {f_1}\ \mathsf d\mu  + \int_E {f_2}\ \mathsf d\mu .$$ 
Let \begin{align}\int_E {f_1}d\mu  &= \sup_{0 \le h \le {f_1}} {I_E}\left( h \right),\\ \int_E {f_2}d\mu  &= \sup_{0 \le g \le {f_2}} {I_E}\left( g \right).\end{align}
Define sets \begin{align}
A &= \left\{ {{I_E}\left( {u} \right):0 \le u \le {f_1} + {f_2}} \right\},\\ B &= \left\{ {{I_E}\left( h \right) + {I_E}\left( g \right):0 \le h \le {f_1},\;0 \le g \le {f_2}} \right\}\end{align}
It is obvious that $B \subseteq A$ since $$0 \le h+g\le f_1+f_2$$ for any $h,g$ and $$I_E(g+h)=I_E(g)+I_E(h).$$
Thus we have \begin{align}\sup B \le \sup A \Rightarrow \sup B &= \mathop {\sup }\limits_{0 \le h \le {f_1},0 \le g \le {f_2}} \left[ {{I_E}\left( h \right) + {I_E}\left( g \right)} \right]\\ &= \sup\limits_{0 \le h \le {f_1}} {I_E}\left( h \right) + \sup\limits_{0 \le g \le {f_2}} {I_E}\left( g \right)\\ &\le \sup\limits_{0 \le u \le {f_1} + {f_2}} {I_E}\left( {u} \right)\end{align}
But I have difficulty proving the opposite inequality. If I can prove every $u$ can be written as $u=h+g$ for some $h,g$, then the prove is done. But it seems it is hard to prove "every $u$ can be written as $u=h+g$". There could be another way around.
Thank you!
 A: As has been pointed out, the problem itself has an easy solution using the Monotone Convergence Theorem. It seems wrong that just additivity for the integral should require such a thing, but that's the way it is.
About your comment regarding whether every $u$ is equal to $h+g$: You're exactly right that that's what we need for a simple direct proof. But you shouldn't feel bad that you can't prove this because it's not true.
Say $E=[0,1]$. Let $f_1(x)=x$ and $f_2(x)=1-x$. Then $u=1$ is a possible $u$, but it's not hard to see that there do not exist $h$ and $g$ with $h+g=1$.
A: Given that the result holds for simple functions, let $f_1$, $f_2$ be nonnegative measurable functions and $\{\varphi_n\}$, $\{\psi_n\}$ be sequences of simple functions such that $\varphi_n\uparrow f_1$ and $\psi_n\uparrow f_2$. Then it is clear that $$\varphi_n+\psi_n\uparrow f_1+f_2$$ and similarly $$ (\varphi_n+\psi_n)\chi_E \uparrow (f_1+f_2)\chi_E$$ for any measurable $E$, so by monotone convergence,
\begin{align}
\int_E(f_1+f_2)\ \mathsf d\mu &= \int \lim_{n\to\infty} (\varphi_n+\psi_n)\chi_E \ \mathsf d\mu\\
&= \lim_{n\to\infty} \int (\varphi_n+\psi_n)\chi_E \ \mathsf d\mu\\
&= \lim_{n\to\infty} \left(\int \varphi_n\chi_E \ \mathsf d\mu + \int \psi_n\chi_E \ \mathsf d\mu \right)\\
&= \int f_1\chi_E \ \mathsf d\mu + \int f_2\chi_E \ \mathsf d\mu\\
&= \int_E f_1\ \mathsf d\mu + \int_E f_2 \ \mathsf d\mu.
\end{align}
