Talk:Lp Space is Subset of Space of Real-Valued Measurable Functions Identified by A.E. Equality

I'm not sure what the confusion is, $\subseteq$ has its standard meaning and the equality is understood as the equality of sets. Basically, the intention is to show that given a function $f$ that is $p$-integrable, any function that is A.E. equal to it is also $p$-integrable, so equivalence classes do not shrink when you change the ambient space the ambient space to $\LL^p$. It is a very minor technical quibble but one that is necessary to address. Caliburn (talk) 17:43, 9 June 2022 (UTC)

(Sorry for edit conflicts) - specifically this result is required for Lp Space is Vector Space, where I want to say that $L^p$ is a vector subspace of $\map {\mathcal M} {X, \Sigma, \R}/\sim_\mu$. That $L^p$ is even a subset is not obvious for the reason above, since they are quotient sets under the same equivalence, but with different ambient sets. We show that by cutting out elements we have only cut out whole equivalence classes, not parts of them. I hope I'm making sense. Caliburn (talk) 17:55, 9 June 2022 (UTC)

I know what you are trying to say, but that inclusion and that equation are formally wrong. Let us see the latter, for example. By definition:

$\eqclass f {\sim_\mu} = \set {g\in\map {L^p} {X, \Sigma, \mu} : f=g \; \text{$\mu$-a.e. }}$

$\eqclass f {\sim_\mu}^\ast = \set {g\in\map {\mathcal M} {X, \Sigma, \R} : f=g \; \text{$\mu$-a.e. }}$

These are not equal as sets, formally.--Usagiop (talk) 18:22, 9 June 2022 (UTC)

However, everything is of course essentially correct. It is similar to saying $\R^2\subseteq \R^3$. What you mean is I guess something like $\R^2\hookrightarrow \R^3$--Usagiop (talk) 18:27, 9 June 2022 (UTC)

I'm not seeing the problem. Were they not equal as sets, there would exist some real-valued measurable function $g$ that is equal to $f$ almost everywhere but is not $p$-integrable, and that's impossible. (as shown in this proof) $\map {\LL^p} {X, \Sigma, \mu}$ here is just defined as the real valued measurable functions that are $p$-integrable, so is a subset of $\map {\mathcal M} {X, \Sigma, \R}$. I might be missing something obvious. Caliburn (talk) 19:03, 9 June 2022 (UTC)

Oh and I realise I didn't prove the case $p = \infty$, I will do that in a little bit. Caliburn (talk) 19:11, 9 June 2022 (UTC)

Please give me the definition of the equality.--Usagiop (talk) 19:12, 9 June 2022 (UTC)

$\eqclass f {\sim_\mu} = \eqclass f {\sim_\mu}^\ast$

That the elements of $S_1 = \set {g\in\map {\LL^p} {X, \Sigma, \mu} : f=g \; \text{$\mu$-a.e. }}$ are precisely the elements of $S_2 = \set {g\in\map {\mathcal M} {X, \Sigma, \R} : f=g \; \text{$\mu$-a.e. }}$. That $S_1 \subseteq S_2$ is immediate from our construction of $\LL^p$ (they do not contain different types of object, they both contain functions $X \to \R$) and this page is intended to prove that there's nothing in $S_2$ that isn't also contained in $S_1$ which is the less trivial implication. Caliburn (talk) 19:18, 9 June 2022 (UTC)

Yes, $S_1 \subseteq S_2$ is true. But why can you say $S_1 = S_2$? Please give me a mathematical definition for this.--Usagiop (talk) 19:23, 9 June 2022 (UTC)

Because if two real valued (measurable) functions are equal A.E., then their ($p$-)integrals (or supremum norm) are the same? So a function that is not $p$-integrable cannot be A.E. equal to a function that is. Could you give an example of an $f$ such that $S_1 \ne S_2$, it might help me understand where the confusion is arising. Caliburn (talk) 19:26, 9 June 2022 (UTC)

OK, sorry I misunderstood your answer. You are right, $S_1=S_2$ is true as sets. So, the inclusion:

$\map {L^p} {X, \Sigma, \mu} \subseteq \map {\mathcal M} {X, \Sigma, \R}/\sim_\mu$

should also be understood as an inclusion of sets of sets by identifying the cosets from left hand side and the cosets from right hand side as abstract sets. Usually, if you talk about cosets, the base set does matter. Since left hand side and left hand side have different base sets, it does not make sense to compare cosets from left hand side with cosets from left hand side. But if you interpret all cosets just as some sets of sets and compare them, then that inclusion and that equality are certainly correct.--Usagiop (talk) 19:43, 9 June 2022 (UTC)

Yes, it feels a bit "exploit"y and more of an unintended thing. Feel free to rework it in a way that makes more sense, if you were confused I'm sure many others will be too. Also feel free to rework any of the other $L^p$ stuff, I really just wanted pages like this up so I could then get onto conditional expectation and Radon-Nikodym derivatives. (though I'll have to do a bit more work for those) Caliburn (talk) 19:49, 9 June 2022 (UTC)

I was confused but you were right. In ProofWiki an equivalence classe is explicitly defined as a set and a quotient set therefore explicitly as a set of sets. So, your statements are also formally correct.--Usagiop (talk) 19:58, 9 June 2022 (UTC)