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Consider the continuous linear functionals $\ell_n, \; n\geq 0$ defined on $C([0,1])$ by $$ \ell_0(f) = \int_0^1 f(t)dt, \;\; \ell_n(f) = \frac{1}{n} \sum_{k = 0}^{n-1} f(\frac{k}{n}), n\geq 1, \;\; f\in C([0,1])$$

Show that $(\ell_n)$ converges weak* to $\ell_0$. Basic Riemann integration can be used without proof.

I have not read any measure theory, can someone please give me an simple proof for this? for me it seems "obvious" that the $\lim_{n\rightarrow \infty}|\ell_0(f) - \ell_n(f)| \rightarrow 0$. But what theorems should be used? Indicator functions and some kind of point wise approximation by constant functions? I've seen similar techniques before.

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The definition of Riemann integrability (in terms of Riemann sums) immediately gives the result. – David Mitra Dec 29 '12 at 18:33
up vote 2 down vote accepted

Let $f \in C([0,1])$. Then by the Fundamental Theorem of Calculus $f$ is Riemann Integrable. Now, using the definition of Riemann Integrability, you get that for each $\epsilon >0$ there exists some $N_\epsilon$ so that for all $n>N_\epsilon$ we have

$$ \left| l_n(f)- l_0(f) \right| < \epsilon \,. (*)$$

This proves that

$$\lim_n l_n(f) = l_0(f) \,;\, \forall f \in C([0,1])$$

Thus $l_n \to l_0$ in the weak* topology.

Note (*) The reason why this holds is the following: $l_n(f)$ is just a Riemann sum for $f$ with the partition $x_0=0<1/n < 2/n<..<n/n=x_n$ and sample points the left end points of the intervals. You know that $\Delta = \frac{1}{n}$.

The Definition of RI says that for each $\epsilon$ there exists a $\delta$ so that for any partition with $\Delta < \delta$ and any sample points we have

$$|l_0(f) - RS( f, P) | < \epsilon \,.$$

Pick some $N_\epsilon > \frac{1}{\delta}$ and then

$$\frac{1}{n} < \frac{1}{N_\epsilon< \delta \,.$$

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