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

• Euklidian-Space
In summary, the problem asks to show that given a sequence of functions ##f_n## converging uniformly to ##f## on a closed interval [a,b] and each ##f_n## being integrable on [a,b], there exists a partition ##P## and a natural number ##N## such that the difference between the lower sums of ##f_n## and ##f## is less than any desired ##\epsilon > 0##. Using the fact that the infimums of the functions are bounded by ##\epsilon## due to the uniform convergence, this can be achieved by choosing an appropriate partition and natural number.
Euklidian-Space

## Homework Statement

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##.

## 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

Euklidian-Space said:

## Homework Statement

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##.

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

Mark44 said:
Can you refresh our memory as to what L in L(f, P) represents? Is it Lebesgue integral?
oh my bad. L(f,p) is the lower sum of the Reimann integral

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

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

Last edited:
Zondrina said:
Just a thought, use the triangle inequality, and say the sum is ##< \frac{\varepsilon}{2} + \frac{\varepsilon}{2}## using the information given.
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?

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

micromass said:
If you can guarantee that ##|f_n(x) - f(x)|<\varepsilon## for all ##x##, does this imply some bound on ##|m_k - m_k'|##?

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

## 1. What does the notation ##\left|L(f_{n}, P) - L(f,P)\right| < \epsilon## mean?

The notation ##\left|L(f_{n}, P) - L(f,P)\right| < \epsilon## is used to represent the difference between the upper and lower Darboux sums of a function ##f## over a partition ##P##. It is a way of measuring the closeness of the upper and lower sums, with a smaller value indicating a closer approximation.

## 2. Why is it important to have a small value for ##\left|L(f_{n}, P) - L(f,P)\right| < \epsilon##?

The value of ##\left|L(f_{n}, P) - L(f,P)\right|## represents the error or difference between the actual value of the integral and the approximation using the partition ##P##. A smaller value indicates a more accurate approximation, which is important in many applications of integration such as calculating areas or volumes.

## 3. How is the value of ##\left|L(f_{n}, P) - L(f,P)\right|## affected by the choice of partition ##P##?

The value of ##\left|L(f_{n}, P) - L(f,P)\right|## is dependent on the choice of partition ##P##, as it represents the difference between the upper and lower sums of the function over that specific partition. Generally, a finer partition (i.e. with smaller subintervals) will result in a smaller value for ##\left|L(f_{n}, P) - L(f,P)\right|##, as it provides a more accurate approximation of the function.

## 4. Can ##\left|L(f_{n}, P) - L(f,P)\right|## ever be equal to ##\epsilon##?

In theory, it is possible for ##\left|L(f_{n}, P) - L(f,P)\right|## to be equal to ##\epsilon##, but this would require a very specific choice of partition and function. In practice, the goal is to choose a partition that results in a value of ##\left|L(f_{n}, P) - L(f,P)\right|## that is smaller than ##\epsilon##, as this indicates a closer approximation.

## 5. How is the concept of ##\left|L(f_{n}, P) - L(f,P)\right| < \epsilon## used in calculus and real-world applications?

The concept of ##\left|L(f_{n}, P) - L(f,P)\right| < \epsilon## is a fundamental part of integration, as it allows us to calculate the value of an integral by approximating it with upper and lower sums. This is useful in many real-world applications, such as calculating the area under a curve, the volume of a solid, or the work done by a variable force. It is also used in theoretical calculus to prove the convergence of sequences and series.

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