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I would like to show the the space of Riemann integrable functions is complete under the uniform metric.

We recall that the uniform metric is: $\rho(f,g) = \sup_{x \in [a,b]}|f-g|$;
and for a metric space to be complete, cauchy sequences must converge.

Take $\{f_n\} \rightarrow f$ Cauchy in uniform metric, we want to show this has a limit. $\forall \epsilon \geq 0, \exists N\ \forall n,m \geq N,\ \sup_{x \in [a,b]}|f_n - f_m| < \epsilon$ For each $x$; $\{f_n\}$ has some limit $y$, we set $f(x) = y$ WTS: $\sup_{x \in [a,b]}|f_n - f| \rightarrow_{n \rightarrow \infty} 0$ For any $x$ Take $m > N s.t. |f(x) - f_m(x)|$, then for all $n \geq N: |f_n(x) - f(x)| \leq |f_n(x) - f_m(x)| + |f_m(x) - f(x)| < 2\epsilon$, therefore: $\forall n \geq N \sup_{x \in [a,b]}|f_n - f| <2\epsilon$, so $f_n$ converges to f uniformly. WTS $f$ is Riemann integrable, fix some $\epsilon > 0$ and let $I(f,P,X)$ where P is partition, and X an evaluation sequence. WTS: $|U(f,P) - U(f_n,P)| < \epsilon < (b-a)2\epsilon$ when $n \geq N$.
$P = \{x_0 < x_1 < \cdots < x_n \}$ $M_i = \sup_{x \in [x_{i-1}, x_i]}f(x)$ , $U(f,P) = \sum_{i = 1}^k M_i(x_{i-1}-x_i)$ and $M_i^n = \sup_{x \in [x_{i-1}, x_i]}f_n(x)$ , $U(f_n,P) = \sum_{i = 1}^k M_i^n(x_{i-1}-x_i)$ $|U(f,P) - U(f_n,P)| \leq \max_i(M_i - M^n_i)(b-a)$

I do not see how to proceed from here, or am I even on the right track? I have the impression i need to use that for $\operatorname{mesh}(P) < \delta, |U(f,P) - L(f,P)| < \epsilon$. Any hints?

Michael Hardy
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rannoudanames
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2 Answers2

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You have already shown that $f_n\to f$ uniformly on $[a,b]$, so for a given $\epsilon>0$ there is an integer $N$ such that

$n>N\Rightarrow \sup |f(x)-f_n(x)|<\epsilon.$ But then

$\int (f_n-\epsilon)\le \underline \int f\le \overline \int f\le \int (f_n+\epsilon)\Rightarrow \overline \int f-\underline \int f\le 2\epsilon (b-a), $

which shows that $f$ is Riemann integrable.

Matematleta
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Sorry, I don't have enough reputation to comment. I don't think your proof of $f_{n} \rightarrow f$ uniformly is correct. How do you know that $N$ is not a function of x? because if it is, then you cannot use the same small $n$ that you got from one $x$ and claim that it works for any $x$.

lucusk
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