Possible error in Folland real analysis proof. I don't think this proof is correct.

Folland defines $L^+$ to be the space of all measurable functions from $X$ to $[0, \infty]$ where $X$ is a measure space.  But then $f - f \chi_E$ is not defined on $X$, since we may have $\infty - \infty$.  
By a.e. he means $f_n(x) \to f(x)$ for all $x \in E$ where $\mu(X \setminus E) = 0$.
So is this a problem?  If so, how can this proof be corrected?
 A: Let's fix this, then. Since $1-\chi_E = \chi_{E^c}$, we can read $f-f\chi_E$ as $f\chi_{E^c}$. So, $f=f\chi_E+f\chi_{E^c}$ which implies
$$
\int f=\int f\chi_E+ \int f\chi_{E^c} = \lim \int f_n\chi_E + 0 = \lim \int f_n
$$
A: $f-f\chi_E = f(1-\chi_E)$, which is $0\cdot \infty = 0 $ when $f(x) = f\chi_E(x)= \infty$
A: There is no error if one keeps in mind the convention which is implicitly used: In order to apply Proposition 2.16, Folland defines the difference of two nonnegative, possibly infinite, measurable functions to be zero at points where both functions are infinite, in accordance with Exercise 2.b.
In order to show that $\int f_n\chi_E=\int f_n$, using the tools he has developed so far, Folland uses the fact that $\int(f_n-f_n\chi_E)=0$ (so that $\int f_n=\int(f_n-f_n\chi_E)+\int f_n\chi_E=\int f_n\chi_E$), which is derived from Proposition 2.16 assuming that $f_n-f_n\chi_E$ is an everywhere defined, nonnegative, possibly infinite measurable function such that $f_n-f_n\chi_E=0$ a.e. (Of course, exactly the same remark applies to $\int f\chi_E=\int f$.)
Regarding the first equality in GDAL's answer, note that the distributive law, $(a-b)c=ac-bc$, does not make sense in the extended reals if, say, $a=b=1$, $c=\infty$, because the right side, $\infty-\infty$ is not defined. If $\infty-\infty=1\cdot\infty-1\cdot\infty=(1-1)\cdot\infty=0\cdot\infty=0$ made sense, then, say, $\infty-\infty=2\cdot\infty-1\cdot\infty=(2-1)\cdot\infty=1\cdot\infty=\infty$ would also make sense and they would be inconsistent. It is for the purpose of proving this corollary that the value $0$ is assigned to $\infty-\infty$, when the latter arises as $f-f\chi_E$ or $f_n-f_n\chi_E$. In general, however, assigning any extended real to $\infty-\infty$ would suffice to ensure that $f-f\chi_E$ and $f_n-f_n\chi_E$ are everywhere defined measurable functions, as in Execise 2.b. 
