real analysis - Question regarding Lebesgue Integrability in $sigma$ -finite spaces

I'm taking a course in measure theory and we defined integrability in a $\sigma$
-finite space as follows: Suppose $\left(X,\mathcal{F},\mu\right)$
is a $\sigma$-finite measure space, a measurable function $f:X\to\mathbb{R}$
is said to be integrable on $X$ (denoted $f\in L^{1}\left(X,\mathcal{F},\mu\right)$) if for every collection $\left\{ X_{m}\right\} _{m=1}^{\infty}$
such that $X_{m}\uparrow X$
, $X_{m}\in\mathcal{F}$
and $\mu\left(X_{m}\right)<\infty$

the following apply:

1. $f$
   is integrable on every set $A\subseteq X$
   such that $\mu\left(A\right)<\infty$
   .




2. The limit $\lim\limits _{m\to\infty}\int_{X_{m}}\left|f\right|d\mu$
   exists and does not depend on the choice of $\left\{ X_{m}\right\} _{m=1}^{\infty}$
   .





3. The limit $\lim\limits _{m\to\infty}\int_{X_{m}}fd\mu$
   does not depend on the choice of $\left\{ X_{m}\right\} _{m=1}^{\infty}$
   .

If said conditions apply then we define $\int_{X}fd\mu=\lim\limits _{m\to\infty}\int_{X_{m}}\left|f\right|d\mu$

Now suppose $\mathcal{G}\subseteq\mathcal{F}$
is a $\sigma$
-algebra on $X$

. Let $f:X\to\mathbb{R}$
be a $\mathcal{G}$
-measurable function such that $f\in L^{1}\left(X,\mathcal{G},\mu\right)$
, is $f$
necessarily in $L^{1}\left(X,\mathcal{F},\mu\right)$
?
Obviously $\mathcal{G}$
-measurability implies $\mathcal{F}$
-measurability but what about integrability?

EDIT: It seems the construction of the integral we did is quite unorthodox, I'll elaborate further on the definitions: Suppose $\left(X,\mathcal{F},\mu\right)$ is a measure space and let $A\subseteq X$ be a subset of finite measure. We define a simple function $f:X\to\mathbb{R}$ to be any function taking a countable collection of real values $\left\{ y_{n}\right\} _{n=1}^{\infty}$. Denote $A_{n}=\left\{ x\in A\,|\, f\left(x\right)=y_{n}\right\}$. Assuming $f$ is measurable we say that $f$ is integrable on $A$ if the series ${\sum_{n=1}^{\infty}{\displaystyle y_{n}\mu\left(A_{n}\right)}}$ is absolutely convergent in which case we define:

$$\int_{A}fd\mu={\displaystyle \sum_{n=1}^{\infty}}y_{n}\mu\left(A_{n}\right)$$
Furthermore, given any measurable function $f:X\to\mathbb{R}$ we say $f$ is integrable on $A$ if there is a sequence of simple functions (as defined) which are integrable on $A$ and converging uniformly to $f$ on $A$. In which case we define: $$\int_{A}fd\mu=\lim_{n\to\infty}\int_{A}f_{n}d\mu$$

Thanks in advance.