Hi all, I'm learning for my analysis exam and I encountered a theorem the proof of which I don't fully understand.(adsbygoogle = window.adsbygoogle || []).push({});

Here it is:

Form of lower and upper Riemann integral

[tex]

\mbox{Let } f \mbox{ be bounded on } [a,b] \mbox{ and let } \left\{D_n\right\}_{n=1}^{\infty} \mbox{ be infinite sequence of dividings of interval } [a,b]. \mbox{ Let } \lim_{n \rightarrow \infty} \Vert D_n \Vert = 0. \mbox{ Then \\ }

[/tex]

[tex]

(R) \int_{a}^b f(x)\ dx = \sup_{n} (f, D_n) \mbox{ (lower Riemann integral)}

[/tex]

[tex]

(R) \int_{a}^b f(x)\ dx = \inf_{n} S(f, D_n) \mbox{ (upper Riemann integral)}

[/tex]

[tex]

\mbox{ Where } s(f, D) \mbox{ and } S(f, D) \mbox{ are lower and upper sums.}

[/tex]

Proof:

Let's choose dividing D and [itex]\epsilon > 0[/itex]. It's sufficient to prove, that

[tex]\exists \ n_0 \in \mathbb{N}: s(f, D_{n_0}) > s(f, D) \ - \ \epsilon

[/tex]

Because then

[tex]\forall \ \epsilon > 0 \ \exists \ n_0 \in \mathbb{N}: \sup_{D'} s(f, D') \ge \sup_{n} s(f, D_n) \ge s(f, D) \ - \ \epsilon

[/tex]

We have fixed D, [itex]\epsilon[/itex] and let

[tex]

K = \sup_{x \in [a,b]} |f(x)|

[/tex]

We choose [itex]n_0[/itex] such, that

[tex]

\Vert D_{n_0} \Vert < \frac{\epsilon}{K.\sharp D'}

[/tex], where [itex]\sharp D'[/itex] = number of intervals in D.

Let

[tex]

P = D_{n_0} \cup D.

[/tex]

[tex]

s(f, D) \le s(f, P) = \sum_{I \in P} (\inf_{I} f).|I| [/tex] (I doesn't contain points of D)

[tex] + \sum_{I \in P} (\inf_{I} f).|I| [/tex] (I contains at least 1 point from D)

[tex] \le s(f, D_{n_0}) + K.\Vert D_{n_0} \Vert .\sharp D

[/tex]

[tex] < s(f, D_{n_0}) + \epsilon \ \ \Box[/tex]

Could you please clarify the main idea and individual steps to me?

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# I don't understand this proof

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