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

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

  2. 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?

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  • $\begingroup$ $E \setminus E_0$ is a null set, so every function is measurable on that. $\endgroup$ Jun 28, 2013 at 17:50
  • $\begingroup$ True, but $f+g$ may not even be defined on $E\backslash E_0$. $\endgroup$
    – David
    Jun 28, 2013 at 17:53
  • $\begingroup$ You can arbitrarily extend $f+g$ to $E\setminus E_0$, all extensions give you a measurable function on all of $E$, and all extensions are equal a.e. $\endgroup$ Jun 28, 2013 at 17:55
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    $\begingroup$ Perhaps it's time to start thinking not in individual functions but in equivalence classes of functions modulo "equal a.e.". Takes a bit to get used to, but is very helpful. $\endgroup$ Jun 28, 2013 at 18:01

1 Answer 1

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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.

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  • $\begingroup$ Thanks to everyone for their helpful ideas. I've posted a possible route of solution in my question above. Can folks look at it and tell me if my argument is valid? $\endgroup$
    – David
    Jun 28, 2013 at 23:01

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