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Let $X$ be a Banach space and assume that $a<b$. For a function $f\colon\left [a,b\right]\to X$ we define a generalization of Riemann integral as follows: a point $u\in X$ we call the integral of $f$ if for any $\epsilon>0$ there is a $\delta>0$ such that for $a=t_0<u_1<t_1<u_2<t_2<\cdots<t_{n-1}<u_n<t_n=b$, where $\max\limits_{i\leq n}(t_i-t_{i-1})<\delta$ we have $\displaystyle\lVert u-\sum_{i=1}^{n}(t_i-t_{i-1})f(u_i)\rVert\leq \epsilon.$

One can show that contiuous functions are integrable in the sense of the definition above.

My question is how to determine the integral of the function $f:\left[0,1\right]\rightarrow c_0$ given by $f(x)=\frac{1}{2}\left(x-\frac{1}{2^n}\right)e_n+\frac{1}{4}\left(\frac{1}{2^{n-1}}-x\right)e_{n+1}$, for $x\in\left[\frac{1}{2^n}, \frac{1}{2^{n-1}}\right]$, $n\in\mathbb{N}$ and $f(0)=0$, where $e_n=(0,...,0,1,0,...,0)$ with $1$ being a the $n$-th place?

Notice that $f$ is continuous so it is integrable.

I guess that the integral may be of the form $\sum_{n=1}^{\infty} \frac{1}{2^{n+1}}e_n$ or something like this but can not deal with it, so I will be very grateful for your help.

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up vote 3 down vote accepted

Your function $f$ can be written as the sum of a series $f(x)=\sum_{n\geq1}\chi_{I_n}(x)f(x)$, with $I_n=\left[\frac{1}{2^n}, \frac{1}{2^{n-1}}\right)$ and $\chi_{I_n}$ the characteristic function of $I_n$. This series converges uniformly, and presumably you can show that your Riemann integrals behave well with such series. Then you can interchange sums and integrals.

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Thanks for the hint, I will try to use it – dawid Oct 12 '11 at 18:46

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