2

I'm having troubles proving following statement

$$ \lim_{\varepsilon \to 0^+} \sum_{i=1}^{N(\varepsilon)} f(\xi_i) \chi_{(x_{i-1},x_i)}(x) = f(x) \qquad \mathrm{a.e. on}\ [0,1]. $$

$f(x)$ is a smooth function with compact support on $[0,1]$, $(x_{i-1},x_i)$ are subintervals of equidistant partition of the interval $[0,1]$ with the length $\varepsilon$ and $\xi_i$ is the center of $(x_{i-1},x_i)$.

Could someone give me a hint how to prove this? Would Riemann's definition of the integral help? Could this be true for pointwise convergence? I don't see why it's only almost everywhere.

goofy
  • 21

1 Answers1

0

To solve the problem, we can use the definition of the limit together with uniform continuity: fix $\eta$ an arbitrary small positive number, and take the corresponding $\delta$ in the definition of uniform continuity.

We have to take the limit with respect to sequences of integers, for example $\varepsilon_n\downarrow 0$. Then call $x_{j,n}$ the corresponding $x_j$ for thiis $\varepsilon$. We have pointwise convergence for every $x$ except elements in the countable set $\{x_{j,n},n\geqslant 1\}$. We cannot expect more for example when we take $\varepsilon_n:=2^{-n}$, because we should have $f(2^{-n})=0$ for all $n$.

Davide Giraudo
  • 172,925
  • Thank you, I've been trying to figure this out but I still don't know exactly what to do. $f(x)$ is continuous on compact set thus it is uniformly continuous, i.e. $\forall \eta > 0\ \exists \delta > 0\ \forall x,y \in [0,1] : |x-y|<\delta$ implies $|f(x)-f(y)|<\eta$. Now I assume that $\epsilon_n < \delta$ ? I'm sorry, I don't quite understand what $x_{j,n}$ is. Could you help me a bit more please? – goofy Mar 01 '13 at 08:59
  • Yes, you choose $n$ such that $\varepsilon_n<\delta$. The $x_{j,n}$ are the points $x_i$ associated with $\varepsilon:=\varepsilon_n$. – Davide Giraudo Mar 01 '13 at 20:15