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I've looked around for answers to this question. It seems like perhaps I don't have enough knowledge of functional analysis to figure out the answer (or even understand the answer), but I'm intrigued.

Consider the set of Riemann Integrable functions over a closed real interval [a,b]. I know that there exists a sequence of Riemann Integrable functions who's limit is not Riemann Integrable. However his begs the question, what is the set of limit points of Riemann Integrable functions?

The question arises from my essay on Daniell Integrals. First we define a lattice of real valued functions L such that

\begin{equation*} (f \lor g) := \max(f,g) \end{equation*} \begin{equation*} (f \land g) := \min(f,g) \end{equation*}

We then define the completion of a lattice $L$, denoted by by $L^{*}$, is defined to be the intersection of all sets $L'$ of real valued functions such that

\begin{equation*} L \subset L' \end{equation*} \begin{equation*} f_{n} \in L', f_{n}\uparrow f \implies f \in L' \end{equation*} \begin{equation*} g_{n} \in L', g_{n}\downarrow g \implies g \in L' \end{equation*}

So essentially it is a larger lattice (I have proven it is a lattice) such that we have closure under monotone limits.

share|improve this question
    
Limit as in pointwise limit? –  Pedro Tamaroff Apr 19 at 15:54
    
Yes, pointwise limit. –  user144261 Apr 19 at 15:59

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