Does there exist approximate forms of the Borsuk–Ulam theorem for classes of "almost continuous" functions? People often say that the Borsuk–Ulam theorem implies that there always exist two antipodal points $x$ and $y$ on the equator such temperature at these two points is the same, i.e. $T(x) = T(y)$ where $T$ is the temperature field on the equator. This of course (implicitly) assumes that $T$ is continuous.
Assuming $T$ to be continuous is probably a good approximation of reality. If one would measure simultaneously the temperature on a dense enough set of points on the equator, one would obtain a discrete function that is "almost continuous" in the sense that the difference $T(x_i) - T(x_{i-1})$ of the temperature of two consecutive points is very small.
This got me thinking that what if we take a function $f$ on the 1-dimensional sphere $S$ (the equator) such that it is "almost continuous" up to scale $\varepsilon$, i.e., the set of points of discontinuity $D \subset S$ satisfy
$$
\left| \lim_{x\to y^+} f(x) - \lim_{x \to y^-}f(x) \right| < \varepsilon \quad \forall y \in D\,.
$$
Now are we guaranteed to find antipodal points $x,y \in S$ such that $|f(x)-f(y)| < \xi(\varepsilon)$ for some function $\xi$ tending to $0$ as $\varepsilon \to 0$? Or something similar? What if we expand our class of "almost continuous functions", for example we could look at functions for which
$$
\left|\limsup_{x \to y} f(x) - \liminf_{x \to y} f(y)  \right| < \varepsilon \quad \forall y \in D\,.
$$
And what if we go to higher dimensions?
In other words, if we only assume "almost continuity" (in some sense) of the temperature field, does there still exist two antipodal points on the equator with practically the same temperature?

My idea would be to approximate the "almost continuous" function with a continuous function. If $f$ is almost continuous up to scale $\varepsilon$, find a continuous function $g$ satisfying
$$
\sup_{x \in S} |f(x)-g(x)| < \varepsilon\,.
$$
Then for $g$ apply the Borsuk–Ulam theorem to obtain antipodal points $z_1, z_2$ s.t. $g(z_1) = g(z_2)$. Now 
\begin{align}
|f(z_1) - f(z_2)| &\leq |f(z_1) - g(z_1)| + |g(z_1) - g(z_2)| + |g(z_2) - f(z_2)| \\
& \leq 2 \varepsilon\,,
\end{align}
and we are done. But I'm not sure how generally this approximation may be applied, especially if we try to loosen our notion of "almost continuity".
 A: *

*Let $f:R\to R$ be periodic with period $2\pi$ , such that $f^+(x)=\lim_{y\to x^+}f(y)$ and  $f^-(x)=\lim_{y\to x^-}f(x)$ exist for every $x\in R.$ Let $g(x)=f(x)-f(x+\pi).$ Then $g^+(x)$ and $g^-(x)$ exist for every $x\in R.$

*Let $C$ be the set of $x \in R$ at which $g$ is continuous.
Assertion: $R$ \ $C$  is countable. 

*Now suppose $e>0$ and $|f^+(x)-f^-(x)|<e$ for all $x.$ Then $\inf |g(x)|< e/2.$
Proof:


*By contradiction, suppose $\inf |g(x)|\geq e/2.$   Let $C^+ =\{x\in C:g(x)\geq e/2\}$ and $C^-=\{x\in C : g(x)\leq e/2\}.$  Since $R$ \ $C$ is countable and $g(x+\pi)=-g(x)$ for all $x\in R$ we have $C^+\ne \emptyset \ne C^-,$ and $\overline {C^+} \cup \overline {C^-}=\overline {C^+\cup C^-}=\bar C=R.$ 


*Now for each $x\in C$ let $I(x)$ be an open interval containing $x,$ such that $y\in I(x)\implies  |g(y)- g(x)|<e/2.$ Observe that for $x\in C^+$ we have $y\in I(x)\implies g(y)\geq e/2$ and that for $x\in C^-$ we have $y\in I(x)\implies g(y)\leq -e/2. $


*Let $U=\cup \{I(x):x\in C^+\}$ and $V=\cup \{I(x):x\in C^-\}.$ The crux of the argument is that $\bar U \cap \bar V=\emptyset.$ For if $x\in \bar U \cap \bar V$ then every nbhd $S$ of x contains $y_S , z_S \in (S\cap I(x))$ \ $\{x\}$  with $g(y_S)\leq -e/2$ and $g(z_S)\geq e/2.$


But: (i). If $y_S , z_S \in (-\infty,x)$ for  arbitrarily small $S$ then $g^-(x)$ does not exist. (ii). If $y_S,z_S\in (x,\infty)$ for  arbitrarily small $S$ then $g^+(x)$ does not exist. (iii). If $x$ lies between $y_S$ and $z_S$  for  arbitrarily small $S$ then $|f^+(x)-f^-(x)|\geq e.$


*So $\bar U ,$  $ \bar V $ are disjoint non-empty closed subsets of $R,$ and their union is $R$ union is $R$ (because $\bar U \cup \bar V \supset \overline { C^+}\cup \overline { C^-}=R). $ This contradicts the fact that  $R$ is a connected space.


