Understanding proof of the Riesz Representation Theorem

46 Views Asked by At

I am studying Stanislaw Lojasiewicz book - "An introduction to the Theory of Real Functions" and I do not uderstand few things. I hope you'll help me. Here is what is written:

G is an open set and $\Gamma(G)$ is defined as a class of all continous, non-negative functions on compact space such that $\varphi\leq 1$ and cl$\{x:\varphi(x)\neq 0\}\subset G$,

$\lambda(G)=\sup_{\Gamma(G)}I(\varphi).$

Now let $G_n$ be a sequence of open sets. Let $L<\lambda(\bigcup\limits_{i=1}^{\infty} G_{i})$. Then there exists $\varphi\in\Gamma(\bigcup\limits_{i=1}^{\infty} G_{i})$ such that $L<I(\varphi).$

That's first thing. Why can we define such constant? How can we know it exist? And why existing of constant $L<\lambda(\bigcup\limits_{i=1}^{\infty} G_{i})$ implies fact that $L<I(\varphi).$

Łojasiewicz, Stanisław, An introduction to the theory of real functions. Transl. from the Polish by G. H. Lawden, ed. by A. V. Ferreira, Wiley-Interscience Publication. Chichester (UK) etc.: Wiley. ix, 230 p. \textsterling 24.95 (1988). ZBL0653.26001.

1

There are 1 best solutions below

4
On

Note that the union plays no role in what you are asking.

If you take any number $L<\lambda(G)$, since $\lambda(G)$ is a supremum there has to exist $\ell\in\{I(\varphi):\ \varphi\in\Gamma(G)\}$ with $L<\ell$. It's just by definition of supremum. And $\ell=I(\varphi)$ for some $\varphi\in \Gamma(G)$, so $L<I(\varphi)$.