$\left|L(f_{n}, P) - L(f,P)\right| < \epsilon$

1. Jun 8, 2015

Euklidian-Space

1. The problem statement, all variables and given/known data
Suppose that $f_{n} \rightarrow f$ uniformly on [a,b] and that each $f_{n}$ is integrable on [a,b]. Show that given $\epsilon > 0$, there exists a partition $P$ and a natural number $N$ such that $\left|L(f_{n}, P) - L(f,P)\right| < \epsilon$.

2. Relevant equations

3. The attempt at a solution
I let P be a partition. And
$$m_{k} = inf\{f(x) : x \in [x_{k-1},x_{k}\}$$
$$m_{k}^{'} = inf\{f_{n}(x): x \in [x_{k-1}, x_{k}]\}$$
I am thinking maybe I can say $|m_{k} - m_{k}^{'}| \leq \frac{\epsilon}{b - a}$, but i do not really know how to justify it. formally anyway

2. Jun 8, 2015

Staff: Mentor

Can you refresh our memory as to what L in L(f, P) represents? Is it Lebesgue integral?

3. Jun 8, 2015

Euklidian-Space

oh my bad. L(f,p) is the lower sum of the Reimann integral

$$L(f,P) = \sum_{k = 1}^{n} m_{k} (x_{k} - x_{k - 1})$$

4. Jun 8, 2015

Zondrina

Just a thought, use the triangle inequality, and say the sum is $< \frac{\varepsilon}{2} + \frac{\varepsilon}{2}$ using the information given.

Last edited: Jun 8, 2015
5. Jun 8, 2015

Euklidian-Space

what exactly is less than $\epsilon/2$? I was thinking something like...

$$|L(f_{n},P) - L(f,p)| = |\sum_{k = 1}^{n} m_{k}^{'}(x_{k} - x_{k - 1}) - \sum_{k = 1}^{n} m_{k} (x_{k} - x_{k - 1}|$$

we then have
$$\sum_{k = 1}^{n} m_{k}^{'} - m_{k} (x_{k} - x_{k - 1})$$

now if i can bound $|m_{k}^{'} - m_{k}$ by epsilon some how then i think that would work?

6. Jun 8, 2015

micromass

Staff Emeritus
If you can guarantee that $|f_n(x) - f(x)|<\varepsilon$ for all $x$, does this imply some bound on $|m_k - m_k'|$?

7. Jun 8, 2015

Euklidian-Space

well since $m_{k}$ and $m_{k}^{'}$ are the infs they are definitely bounded by $\epsilon$ if $|f_n(x) - f(x)|<\varepsilon$ for all $x$; which is true because we have uniform convergence