- #1
twoflower
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Hi all, I'm learning for my analysis exam and I encountered a theorem the proof of which I don't fully understand.
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?
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?