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A collection $\{f_n\}$ of real valued functions is said to be HK-equi-integrable on $I=[a,b]$, if there exists a gauge $\delta$ on $I$ such that for every $\epsilon>0$, there exists a $\delta$-fine tagged partition $\mathcal{P}$ of $I$ such that $\|S(f,\mathcal{P})-\int_I f\|_X<\epsilon$ for every $n$. That is, the gauge $\delta$ is independent on $n$.
Where gauge $\delta$ is any positive function $\delta:[a,b]\rightarrow\mathbb{R}^+$. And $S(f,\mathcal{P})=\sum_i f(\xi_i)(x_i-x_{i-1})$.

Now my question is that does equi-integrability of $\{f_n\}$ and point wise convergence of $\{f_n\}$ to $f$ implies uniform convergence of the sequence $\{f_n\}$?

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  • $\begingroup$ Please confirm: 1) The $f_n$ are assumed Riemann integrable. 2) Here the norm $\|\cdot\|_X$ is just the absolute value. 3) Here $\delta$-fine means $x_{i+1}-x_i\le \delta(\epsilon)$ for $i=0,\dots, n-1$. $\endgroup$
    – Pietro Majer
    Commented Aug 9, 2011 at 11:37
  • $\begingroup$ This question is a duplicate (slightly expanded) of mathoverflow.net/questions/72447 . @Pietro: The whole question is about Henstock–Kurzweil integral. What would be the point of asking $f_n$ to be Riemann integrable? $\endgroup$
    – Emil Jeřábek
    Commented Aug 9, 2011 at 12:06
  • $\begingroup$ no special point, but to have a clear question. $\endgroup$
    – Pietro Majer
    Commented Aug 9, 2011 at 13:23
  • $\begingroup$ Oh I didn't notice all these clones of the question. $\endgroup$
    – Pietro Majer
    Commented Aug 9, 2011 at 13:26
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    $\begingroup$ In the closed version of the question, Emil Jeřábek says: ... I think that the intended definition of HK-equi-integrability should read: for every $\epsilon>0$ there exists a gauge $\delta$ such that for every $n$ and every $\delta$-fine partition $\mathcal P$, $|S(f_n,\mathcal P)−\int f_n|\le \epsilon$. $\endgroup$
    – Gerald Edgar
    Commented Aug 9, 2011 at 14:24

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What if you try these functions $f_n$ on $[0,1]$ ... piecewise linear, with $f_n(0) = 0$, $f_n(1/n)=-1$, $f_n(2/n)=1$, $f_n(3/n)=0$, $f_n(1)=0$.

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