$f, g$ measurable function on $E$ that are finite a.e. on $E$ Reading Royden 4th Ed, page 56. I've just finished proving Proposition 5: Let $f$ be an extended real-valued function on $E$.


*

*If $f$ is measurable on $E$ and $f=g$ a.e. on $E$, then $g$ is measurable on $E$.

*For a measurable subset $D$ of $E$, $f$ is measurable on $E$ if and only if the restrictions of $f$ to $D$ and $E\backslash D$ are measurable.
I was able to prove this proposition. However, the difficulty begins for me in the ensuing paragraph:
"The sum $f+g$ of two measurable extended real-valued functions $f$ and $g$ is not properly defined at points at which $f$ and $g$ take infinite values of opposite sign. Assume $f$ and $g$ are finite a.e. on $E$. Define $E_0$ to be the set of points in $E$ at which both $f$ and $g$ are finite."
OK, I'm good so far. Understand that completely. Continuing with the paragraph ...
"If the restriction of $f+g$ to $E_0$ is measurable, then, by the preceding proposition, any extension of $f+g$, as an extended real-valued function, to all of $E$ is also measurable."
I am in trouble with the last statement, as $f+g$ might not be defined on $E\backslash E_0$.
Then Royden continues with Theorem 6: Let $f$ and $g$ be measurable functionis on $E$ that are finite a.e. on $E$. Then $f+g$ is measurable on $E$.
The first line of the proof is as follows:
Proof: By the above remarks, we may assume $f$ and $g$ are finite on all of $E$.
This last line just blows me away. I don't get it.
Any thoughts that might be helpful?
Possible Solution
I want to thank folks for their suggestions. I'd like to post a possible interpretation that will help solve my difficulties. Please let me know if I am finally thinking properly.
We start by letting $f$ and $g$ be measureable functions on $E$ that are finite a.e. on $E$. Let $E_0$ be the set of points in $E$ at which both $f$ and $g$ are finite. Thus, $m^*(E_0)=0$ and that makes $E_0$ a measurable set. Because both $f$ and $g$ are measurable functions on $E$, that make $E$ a measurable set. Because the set of measurable sets is a $\sigma$-algebra, $E-E_0=E\cap E_0^C$ is also a measurable set. I will now define two new functions, extensions of the given $f$ and $g$ to all of the set $E$.
$$f^*(x)=\begin{cases}
f(x), & \text{if $x\in E_0$}\\
0, & \text{otherwise}
\end{cases}$$
And:
$$g^*(x)=\begin{cases}
g(x), & \text{if $x\in E_0$}\\
0, & \text{otherwise}
\end{cases}$$
The first thing to note is that $f=f^*$ a.e. on $E$, so $f^*$ is measurable on $E$. The second thing to note is that $g=g^*$ a.e. on $E$, so $g^*$ is measurable on $E$. This is by part (1) of the opening proposition above.
Next, both $f^*$ and $g^*$ are finite on all of $E$. Now, suppose that I succeed in showing that $f^*+g^*$ is measurable on $E$. The next thing to note is that $f+g$=$f^*+g^*$ almost everywhere on $E$. Again, by part (1) of the opening proposition above, $f+g$ is measurable on $E$.
I think I have it. Am I correct in my thinking?
 A: There is an intermediate step there which is not explicitly stated. Assume that $f$ and $g$ are defined on all of $E$ and that they are finite a.e.
Then $f+g$ is not necessarily defined on all of $E$. But it isn't necessarily defined only on $E_0$ either (i.e. the set where both $f$ and $g$ are finite). Rather, $f+g$ is defined on something in the middle: the set of all points such that $f$ and $g$ are not both infinite and of opposite sign.
Call this set $E_1$. Then the proposition you quoted at the top says that $f+g$ is measurable on $E_1$ since the restrictions of $f+g$ to $E_0$ and $E_1 - E_0$ are measurable. It does not actually say anything about all of $E$ because, as you noted, $f+g$ is not necessarily defined on all of $E$.
But then, as Daniel Fischer pointed out in the comments, $f+g$ can be arbitrarily extended to all of $E$ such that the extension is measurable on all of $E$. Hence wording in Royden that you quoted:

If the restriction of $f+g$ to $E_0$ is measurable, then, by the
  preceding proposition, any extension of $f+g$, as an extended
  real-valued function, to all of $E$ is also measurable.

