Showing well-definedness of an equivalence class in $L^{\infty}(\mathbb R)$

Question

Let $\mathbb L:=\{ f\in L^{\infty}(\mathbb R): \operatorname{there exists a representative}$ $\widetilde{f}$ so that $\lim\limits_{x \to \infty}f(x):=\lim\limits_{x \to \infty}\widetilde{f}(x)\operatorname{exists}\}$

Show that the $\mathbb L$ is well-defined.

My idea:

Let $f \in \mathbb L$ arbitrary and $\widetilde{f}, \widetilde{g}$ be representatives thereof, so that $\lim\limits_{x \to \infty} \widetilde{f}(x):=a<\infty$ and $\lim\limits_{x \to \infty} \widetilde{g}(x):=b<\infty$

Let $\epsilon > 0$. We know that by the above definition there exists $x_{0}$ so that for all $x\geq x_{0}$: $\vert\widetilde{g}(x)-b\vert <\frac{\epsilon}{2}$ and $\vert\widetilde{f}(x)-a\vert<\frac{\epsilon}{2}$. Since we are in $L^{\infty}$, we know that $\widetilde{f}=\widetilde{g}$ almost everywhere. And then I get stuck. In the solutions, it states that after a particular $x_{1}$, $\widetilde{g}(x)=\widetilde{f}(x)$ for all $x \geq x_{1}$. Why is this so? Surely, it is only almost everywhere?

Answer 1

Suppose $a \neq b$. Take $\epsilon =\frac {|a-b|} 2$. If $a>b$ then $\widetilde f (x) >a-\frac {\epsilon} 2 >b+\frac {\epsilon} 2 > \widetilde g(x)$ for $x \geq x_0$. This implies that $\widetilde f$ and $\widetilde g$ don't represent the same $f$. Similar argument holds if $a<b$.

Answer 2

As far as I can see, you're proving well-definedness of $\lim_{x\to+\infty}$ on $\mathbb L$, not well-definedness of $\mathbb L$ (which would be trivial).

You're right about "almost everywhere", and you're also right about the general idea of the proof. Notice that what you wrote is true for all $\epsilon>0$. So if $a$ and $b$ were different then you could take $\epsilon<|a-b|$. But then the inequalities you proved about $\epsilon/2$ and the triangle inequality imply that $\tilde f(x)\neq\tilde g(x)$ for all $x\geq x_0$. That's incompatible with $\tilde f$ and $\tilde g$ agreeing almost everywhere.