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Suppose $\{f_n\}_{n\geq1}$ is a series of bounded variation functions on $[a,b]$ satisfying: $$ T_a^b(f_n)\leq M_1, | f_n(a) |\leq M_2 $$ for all $n$, where $M_1, M_2$ are real numbers.

I'm wondering: if $f_n$ converges to a function $f$ everywhere, does it imply

$$ T_a^b(f_n) \rightarrow T_a^b(f)$$

If it doesn't hold, can we make a few justifications?

It just occurred to me when I'm studying real analysis. Although it seems intuitive, I can't directly think of a proof or give a counterexample.

Thanks!

R. Feng
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1 Answers1

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For simplicity, assume that $a=0$ and $b=1$. We define $f_n$ as the indicator function of the interval $(0,1/n)$. In this way,

  • $T_0^1\left(f_n\right)=2$ (reached with the subdivision $\{0,1/(2n),1\}$ and no other subdivision can do better),
  • $f_n(0)=0$ for all $n$ and
  • $f_n(x)\to 0$ for all $x\in[0,1]$.

However, under pointwise convergence, it is true that $T_a^b(f)\leqslant \liminf_{n\to \infty} T_a^b (f_n)$.

Davide Giraudo
  • 172,925