This often only requires measur-ability but that has been weakened to baire measurability (1) I think. Or if F is has bounded above on a set of positive measure. Or just having an unique local upper bound  in some cases. I know that monotone functions are often borel measurable. I am not sure about strict monotone functions

This article says that '(strict quasi  /or rather semi strict quasi convexity, in some more modern terminology) plus midpoint con-vextiy implies convexity',

However I believe that 'quasi convexity and their concave counterparts, as well as their strict and semi-strict counterparts are all implied, or if the function is real valued, and striclty monotone increasing from a real interval to a real interval

So would strict monotone increasing give the same result?

i note that in the article it does not restrict the co-domain to be a closed interval; it explicitly allows for the domain to be (which was the weakening allowed for by strict quasi convexity)

It also says says the domain, which may be a closed and bounded real interval such [0,1], must be also be a convex subset of not Necessarily normed (general) real linear space.

So I am not sure exactly that means with regard to the other properties the function must have.

The restriction of F to the interval [0,1) must be continuous and convex. F need not to be continuous at the point 1.

Dear Roman, I was about to send you through some material.

So are you saying that that the strict monotonicity, given midpoint convexity and F(0)=0 and F(1) will the convexity of this midpoint convex function; or close enough, . OTherwise something would have to be done about the end points (i note that even convex functions have some issues with continuity at the end points). A lot of the time, the Function only has to be continuous at a point.

I presume that strictly quasi convexity and strict quasi concavity is already implied? Would adding F(0.5)=0.5 change anything?

Ipresume some of the issues, might be that there could be a turn in ensuring that the functio meets both end-points. The function is allowed to be concave as well. Will spec

What ensure this degree of continuity- the strict quasi convexity, given F is strictly monotonic increasing plus  Midpoint convexity, with F(0)=0 and F(1)=1. I presume some of the issues can arise if there are more then two fixed points.,

@roman ger does the continuity of midpoint convex functions to the restriction (0.1) under lebesgue measurability, require F strict monotonic increasing or F being  monotone increasing ?

I apologize; it makes sense; I presume that its because one cannot express the limit from below as F(x) for as x approaches 1 from below using( less then or equal in midpoint convexity alone), unless one has something like symmetry (which makes it concave as well as x increases). Although I though that generally if a midpoint convex function was locally bounded at all points in the interior from above and below it was often c onsidered continuous; or that if the limit from above exists and is equal to the defined pt is bounded and the continuous from below as well ie s everwhere (but maybe not from below)

I

I presume that once, the end points and the domain and co-domain is specified , with F(1)=1 and F(0)=0 ( and perhaps  strict mon-otonicity) this  trivially defines the limit from above for 1, and the limit from below for x=0 with F(0)=0 and F(1)=1 with co-domain domain being set to [0,1] and [0,1].

And this holds in general, whether its midpoint convex or not , if Only  wanted to show continuity at the end points .

That is that only  the rhs continuity @ 0;  with F(0)=0 and co-domain being [0,1] is required,  and equals the  function value at 0, which is defined there to be 0 and that the lhs limit exists for x=1 and that F(1)=1 is defined there and equals said rhs limit  is necessary; at least with monotonicity but maybe not even that. Because in some sense it would not make sense to take these limits if the domain is restricted to [0,1] and the (co-domain is as well) to take the LHS limit of x=0 and the rhs limit of 1? One would have to go outside the domain?

As it would not make sense if the domain is equal to [0,1] to require a derivation that both lhs and rhs limits exist and equal each other @0 and @ 1, both fromabove and below for each individuallyt,  and respectively also equal F(1)=1 and F(0)=0.  but just that the right hand limit for 0a nd the left hand limit for 1 do?

I presume that one only needs to show a limit from from below at x=1, F(1)=1, if with f(0)=0 , that is continuous and bounded from above, to show that  F is midpoint convex function is continuous (with or without monotonicity?)

I know that if its bounded from below , and continuous at x=1 ,the entire function domain is locally bounded  above and continuous from  belowe (so perhaps one requires both) if F is real valued midpoint convex on a closed domain;  and that one only then has to how continuity from above; by way for showing that its continuous from below at x=0; given its midpoint convex.

Whether or not the function is midpoint convex; that is, if one only wanted to show continuity as those two points. I presume that monotonically alone or strict monotonic increasing' alone would entail not anything given this (or midpoint convexity is insolation); that is continuity at the end pts, would not otherwise show anything but continuity @ endpoints, unless the function was both monotonic and midpoint convex or had some other property. Otherwise, the timit from above or below must be shown for the interior points (if one did not have midpoint convexity or monotonicity but only one or neither properties); or am I correct that one requires this to say that the limit from above always matches the defined pt if the limit from below does (even for the end points)

with the function being strictly monotonic,t ( there is limit from below for zero or limit from above for one to make sense)

I presume that this would hold regardless of midpoint convexity or monotonicity, but that it would not have any implications for the interior pts, unless one had both properties? When the function is monotonic, I presume it rules out singularities and not just' Removable discontinuities';

ie what about if F its not midpoint convex but the function is does not uniquely defined at a certain pt but F strictly monotonic;  the lhs and rhs limits exists, then if they equal each other, that mean that the functio value  uniquely defined at that point;

Strict monotonic or even monotonic F,  cant have removable continuities, and only jump discontinuities,  but what about singularities? Does  monotonicity  that deal with the case where the function value at that point is not uniquely defined or unknown (a singularity)

That is where  the two left and rhs limits (exist by monotonicity) and happen to equal each other, but its not that they dont equal the function value at that pt but that undefined in the sense that its unknown. Or is F  just trivially defined at some pt ( if that pt is in the  domain) .

One does not have to uniquely define the function value at that pt before-hand- prior to showing that the limits equal each other (given that they exist as well by monotonicity). As showing that these (already existing) lhs and rhs limits, equal each other, for monotonic  does this. As the function value exists at that pt, even if its not uniquely known a priori if that pt is the domain (by definition of a function). So do singularities only pertain to cases where the x value is not in the domain of the function to begin with at all; as opposes to when it is i the domain, and is defined, but not known or uniquely specified, as to what said value is?.

Ie if a midpoint convex function can be shown to continuous at 0 and 1 with F(0)=0 and F(1)=1 and its not necessarily monotonic, or stricly, 

or if a strictly monotonic function can be show to be continuous at both end pts, but not midpoint point convex; it would not have any implications for the continuity of the interior pts I presume? With those properties in isolation. Likewise for midpoint convexity by itself

(I presume that  strict monotonicity would be required if the co-domain was arbitrary, or perhaps not)

And that the limit from above  as x approaches zero. That is if  lim F(x) to 0 (from above) F(x)\to 0 from above as F(0)=0. I presume this always the case.?at least for the end points. In the case of a monotonic function that is midpoint convex; then that will entail continuity (but with midpoint convexity alone, or strict monotonicity alone it would not.  I presume neither by themselves, (with continuity at 0 and 1, without both midpt convexity and monotonicity, entail anything about the interior?

@roman Ger, Is it correct that if the function is midpoint concave instead of midpoint convex,, that continuity at 0 would bean issue .Itt would be continuous @ 1 but instead would have problems at 0; the opposite of the midpt convex case .

That is of F to the interval (0,1] must be continuous and concave x.  but F need not to be continuous at the point 0, if midpoint convexity in the above question was replaced with midpoint concavity?