Can someone give a simpler solution to this proof? If $f $ is a measurable function and $f=g $ almost always, then $ g$ is a measurable function. Can someone give a simpler solution to this proof?
If $f $ is a measurable function and $f=g $ almost always, then $ g$ is a measurable function.
Suppose then first that $f,g: X \rightarrow Y$ where $(X, \mathcal{A}, \mu), (Y, \mathcal{B}, \lambda) $ are measure spaces, where $\mu$ is complete.
Then, if $f=g$ at almost every point, we have that there exists a set of measure null $ E \in \mathcal{A}$ such that $f(x) =g(x)$ for each $x \in E^c$.
So let $B \in \mathcal{B}$, and note that what we want to see is that $ g^{-1}(B) \in \mathcal{A}$ . Now notice that
[center]$g^{-1}(B)=\{x \in X: g(x) \in B\} = \{x \in E^c: g(x) \in B\} \cup \{x \in E: g(x) \in B\} = \{x \in E^c: f(x) \in B\} \cup \{x \in E: g(x) \in B\} = \left( f^{-1}(B) \cap E^c \right) \cup \{x \in E: g(x) \in B\}$ [/center]
Then, since $f^{-1}(B)$ and $E^c$ are measurable (the first set is measurable because $f$ is measurable) their intersection is too. Also, since $\{x \in E: g(x) \in B\} \subset E$, $\mu(E)=0$ and the measure is complete, we have that $\{x \in E: g(x) \in B\}$ is also measurable in $X$. However, we get that
[center]$\left( f^{-1}(B) \cap E^c \right) \cup \{x \in E: g(x) \in B\} \in \mathcal{A }$[/center]
then by the above, that $g^{-1}(B) \in \mathcal{A}$ as we wanted to see.
If the measure space $X$ isn't complete, then this doesn't have to be true, although I can't think of a simple example of it right now. If it occurs to me I edit the message putting it.
we can demonstrate the following:
[i]Let $(X, \mathcal{A}, \mu), (Y, \mathcal{B}, \lambda)$ measure spaces such that $\mathcal{B}$ is not $\sigma$-trivial algebra, then they are equivalent:[/i]
[list type=decimal][li][i]$\mu$ is a full measure.[/i][/li]
[li][i]For any functions $f, g: X \rightarrow Y$ such that $f$ is measurable and $f=g$ in almost everything point, it is true that $g$ is measurable.[/i][/li][/list]
That [b]1[/b] implies [b]2[/b] is what we have proved before, then let's see that [b]2[/b] implies [b]1[/b].
Let $E \in \mathcal{A}$ be such that $\mu(E)=0$ and $A \subset E$, we must see that $A \in \mathcal{A}$.
Consider now $B \in \mathcal{B}$ not empty and distinct from $Y$, and any two distinct points $a, b \in Y$ such that $a \in B$ and $b \not \in B$. Let us define $f:X \rightarrow Y$ such that $f(x)=a$ if $x \in E$ and $f(x)= b$ if $x \in E^c$, and similarly we define $g$ but considering $A$ instead of $E$.
It is then easy to see that $f$ and $g$ are measurable if and only if $E$ and $A$ are measurable sets respectively.
Thus, $f$ is measurable because $E$ is a measurable set and $f=g$ at almost every point because $f(x)=g (x)=b$ for each $x \in E^c$. Therefore, by [b]2.[/b] we have that $g$ is measurable and thus, $A$ must be a measurable set.
 A: Let $(X, \mathcal{A}, \mu) $ complete measure space.
$E^c=\{x\in X:f(x) =g(x)\}$
Given $\mu(E) =0$, thus $E^c$ also measurable.
Let $B\in \mathcal{B}$.
Then $g^{-1}(B) =(f^{-1}(B) \cap E^c)\cup (g^{-1}(B)\cap E) $
$f^{-1}(B) \cap E^c\in \mathcal{A}$ and $g^{-1}(B)\cap E\subset E$ and $\mu(E) =0$.
Hence $g^{-1}(B)\in\mathcal{A}$

Consider $X=\Bbb{R}, \mathcal{A}=\{A\subset \Bbb{R}: A\text{ is countable or } A^c \text{is countable} \}$
and $\mathcal{B}=$Lebesgue sigma algebra.
Let $\mu=\delta_0$ defined on the measurable space $(X, \mathcal{A})$ by $$\mu(A) =\begin{cases} 1 &0\in A \\0&\text{otherwise}\end{cases}$$
Let $\lambda$ is the lebesgue measure on $(X, \mathcal{B}) $
Now consider two functions $f, g $ defined by $f(x) =0$ and $g(x) =x, \forall x\in X$
Then $f$ is clearly measurable and $f=g$ on $\{0\}$ but $g$ is not measurable.
Note:
$\begin{align}&\mu\{x\in X : f(x) \neq g(x) \}\\&=\mu(\Bbb{R}\setminus\{0\})\\&=0\end{align}$
$(0,1)\in\mathcal{B}$ but $g^{-1}(0, 1)=(0, 1) $ neither countable nor has countable compliment. Hence $(0, 1) \notin \mathcal{A}$
