If $\mathcal{G}\subset\mathcal{F}$ are two $\sigma$ algebras on a set $X$, $\mu$ is a nonnegative measure on $(X,\mathcal{F})$ and $f:X\to[0,+\infty]$ is $\mathcal{G}$-measurable, then there are two possible definitions of $\int f~d\mu$. The first is as an $\mathcal{F}$-measurable map:
$$\int_\mathcal{F} f~d\mu:=\sup\left\{\int\varphi~d\mu~\Big|\:\begin{array}{c}0\leq\varphi\leq f\text{ simple}\\\mathcal{F}\text{-measurable map}\end{array}\right\}$$
the second is as a $\mathcal{G}$-measurable map:
$$\int_\mathcal{G} f~d\mu:=\sup\left\{\int\varphi~d\mu~\Big|\:\begin{array}{c}0\leq\varphi\leq f\text{ simple}\\ \mathcal{G}\text{-measurable map}\end{array}\right\}$$
These two definitions coincide as follows for instance from an application of Fatou's lemma (for $\mathcal{F}$-measurable maps) and the dominated convergence theorem (for $\mathcal{G}$-measurable maps).
Question: is there a direct argument proving $\int_\mathcal{G} f~d\mu=\int_\mathcal{F} f~d\mu$ that I'm overlooking? Maybe on the basis of a monotone class argument?
Proof I had in mind: Note that by definition of both integrals the result is true for indicator functions $1_A$, $A\in\mathcal{G}$. Furthermore, by linearity of both $\int_\mathcal{G}\cdot~ d\mu$ and $\int_\mathcal{F}\cdot~ d\mu$, the result holds for all nonnegative $\mathcal{G}$-measurable simple functions ${\color{blue}{(1)}}$. Furthermore, since every $\mathcal{G}$-measurable simple function is a $\mathcal{F}$-measurable simple function, one always has
$$
\int_\mathcal{G} f~d\mu\leq\int_\mathcal{F} f~d\mu.
$$
Thus, if $\int_\mathcal{G} f~d\mu=+\infty$, the result is established. Now suppose $\int_\mathcal{G} f~d\mu<+\infty$. For every $n\in\Bbb{N}$ we set for instance
$$
f_n = \sum_{i=1}^{n2^n}2^{-n}1_{[i/2^n\leq f]}.
$$
Then $0\leq f_n\leq f$ is a nonnegative $\mathcal{G}$-measurable simple function which converges pointwise to $f$. We get that $f\in L^1(\Omega,\mathcal{F},\mu)$ and the reverse inequality since by Fatou's lemma $\color{red}{(2)}$ and dominated convergence $\color{green}{(3)}$
$$\begin{eqnarray}
0
\leq
\int_\mathcal{F} f~d\mu
=
\int_\mathcal{F}\liminf_n f_n~d\mu
& \overset{\color{red}{(2)}}\leq &
\liminf_n\int_\mathcal{F} f_n~d\mu\\
& \overset{\color{blue}{(1)}}= &
\liminf_n\int_\mathcal{G} f_n~d\mu\\
& \overset{\color{green}{(3)}}= &
\int_\mathcal{G} f~d\mu.
\end{eqnarray}$$
Better proof. As noted in @KaviRamaMurthy's answer, the right tool is the monotone convergence theorem. We know by the first point ${\color{blue}{(1)}}$ that both definitions agree on nonnegative simple $\mathcal{G}$-measurable functions. Taking the $0\leq f_n\leq f_{n+1}\leq f$ defined above, one has by two applications of the monotone convergence theorem that
$$
\int_{\mathcal{F}} f~d\mu
\overset{\text{MCT}}=
\lim_n
\int_{\mathcal{F}} f_n~d\mu
\overset{{\color{blue}{(1)}}}=
\lim_n
\int_{\mathcal{G}} f_n~d\mu
\overset{\text{MCT}}=
\int_{\mathcal{G}} f~d\mu
$$
Best Answer
In the books I have used $\int fd\mu$ is defined using a specific construction of simple functions approximating $f$: $f_n(x)=\frac {i-1} {2^{n}}$ if $\frac {i-1} {2^{n}} \leq f <\frac i {2^{n}}$, $i \leq N2^{n}$ and $f_n(x)=N$ for $i >N2^{n}$. And $\int fd\mu=\lim \int f_n d\mu$. In the present case the same sequence is also a sequence of simple functions on $(X, \mathcal G)$ increasing to $f$ so the definition of $\int fd\mu$ is same whether you are using the sigma algebra $\mathcal G$ or $\mathcal F$.