strict convexity with a measure theoretic property Suppose $(x_n)$ is a positive sequence of reals converging to $x$. Furthermore we have a measure space $(E,\mathcal{E},\mu)$ given, with finite measure $\mu$. There are a measurables nonnegative and real valued maps $(f_{x_n})$ and a $f_x$ (depending on $x_n$ and $x$) such that for some $\epsilon >0$ given
$$\lim\sup_n \mu\{y\in E:|f_{x_n}(y)-f_x(y)|>\epsilon\}>\epsilon\tag{1}$$
Moreover
$$\lim\sup_n \mu\{y\in E:|f_{x_n}(y)-f_x(y)|>\epsilon,f_{x_n}(y)+f_x(y)\le \frac{1}{\epsilon}\}>\epsilon\tag{2}$$
Furthermore there is a strictly convex map $H$ given, hence
$$H(\frac{1}{2}(f_{x_n}(y)+f_x(y)))<\frac{1}{2}(H(f_{x_n}(y))+H(f_x(y)))\tag{3}$$
Now we should deduce from $(2)$ and $(3)$ the existence of $\delta>0$ such that
$$\lim\sup_n\mu\{H(\frac{1}{2}(f_{x_n}(y)+f_x(y)))\le\frac{1}{2}(H(f_{x_n}(y))+H(f_x(y)))-\delta\}>\delta$$
It is clear that I can find for very $n$ a $\delta(n)>0$ s.t.
$$H(\frac{1}{2}(f_{x_n}(y)+f_x(y)))\le\frac{1}{2}(H(f_{x_n}(y))+H(f_x(y)))-\delta(n)$$
However I'm struggling with the uniformity in $n$ and the measure of this event.
 A: If $H$ is both strictly midpoint-convex and measurable then there exists a strictly increasing weak derivative $H'$.
Therefore for fixed $y$ the function
$$ g_y(z) = \frac{H(z) + H(f(y))}2 - H\left(\tfrac12 (f(y)+z)\right) $$
Is strictly decreasing and continuous with $g_y(f(y)) = 0$ therefore $g_y$ has a continuous inverse and we may find some $\delta_y>0$ such that $g_y(z)\le\delta_y \Rightarrow \left|z-f(y)\right|\le\epsilon$.
Furthermore as $\mu$ is finite we may choose some $\delta <\frac 12\epsilon$ such that $\mu\{y\in E:\delta_y < \delta\}\leq\frac 12 \epsilon$.
So if $$H(\frac{1}{2}(f_{x_n}(y)+f_x(y)))>\frac{1}{2}(H(f_{x_n}(y))+H(f_x(y)))-\delta$$
we must have $g_y(f_{x_n}(y)) <\delta$ hence either $\left|f_{x_n}(y)-f(y)\right|\leq\varepsilon$ or $\delta_y<\delta$.
That is
$$\begin{align}
\{y\in E:g_y(f_{x_n}(y)) <\delta\} &\subset \{y\in E:\left|f_{x_n}(y)-f(y)\right|\leq\varepsilon\} \cup \{y\in E:\delta_y<\delta\}, \\
\{y\in E:g_y(f_{x_n}(y)) \geq \delta\} &\supset \{y\in E:\left|f_{x_n}(y)-f(y)\right|>\varepsilon\} \setminus \{y\in E:\delta_y<\delta\}. 
\end{align}$$
Therefore for there exist infinitely many $n$ such that
$$\begin{align}
\mu\{y\in E:g_y(f_{x_n}(y)) \geq \delta\} &\geq \mu\{y\in E:\left|f_{x_n}(y)-f(y)\right|>\varepsilon\}  - \mu\{y\in E:\delta_y<\delta\}. \\
&\geq \epsilon - \frac 12\epsilon \\
&>\delta
\end{align}$$
and 
$$\lim\sup_n\mu\{H(\frac{1}{2}(f_{x_n}(y)+f_x(y)))\le\frac{1}{2}(H(f_{x_n}(y))+H(f_x(y)))-\delta\}>\delta$$as required.
