- #1

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- TL;DR Summary
- uniformly to ##f##, where ##f## is some function. Suppose also that each ##f_n## is Riemann-integrable. Show that ##f## is Riemann-integrable also, and that the integral of ##f## is equal to that of ##f_n## whenever ##n## is sufficiently large.

Let ##\epsilon>0##. Choose ##N\in\mathbb{N}## s.t. for each integer ##n## s.t. ##n\geq N##,

$$|\sup\{|f-f_n|(x):x\in D\}|<\frac{\epsilon}{3}$$

where ##D## denotes the intersection of the domains of ##f## and ##f_n##.

Choose a partition ##P:=\{x_0,\ldots,x_m\}## with ##x_i<x_{i+1}## where ##0\leq i\leq m-1## s.t. the following holds.

%%% EDIT: The function to be integrated has been corrected from ##f## to ##f_n##.

$$|f_n(s_i)-f_n(t_i)|\Delta x_i<\frac{\epsilon}{3}$$

$$\Delta x_i\leq 1$$

$$|f_n(t_i)\Delta x_i -\int_{x_i}^{x_{i+1}} f_n|<\frac{\epsilon}{3}$$

where ##s_i,t_i\in[x_i,x_{i+1}]## and ##\Delta x_i:=x_{i+1}-x_i##.

Hence,

$$\eqalign{%

\epsilon>&|f_n(s_i)-f(t_i)|\Delta x_i\cr

+&||f-f_n||+|f_n(s_i)\Delta x_i-\int_{x_i}^{x_{i+1}}f_n|\cr

\geq&|f_n(s_i)-f_n(t_i)|\Delta x_i\cr

+&|f(s_i)-f_n(s_i)|+|f_n(t_i)\Delta x_i - \int_{x_i}^{x_{i+1}}f_n|\cr

\geq&|(f_n(s_i)-f_n(t_i))\Delta x_i\cr

+&(f(s_i)+f_n(s_i))\Delta x_i\cr

+&(f_n(t_i)\Delta x_i-\int_{x_i}^{x_{i+1}} f_n)|\cr

=&|f(s_i)\Delta x_i-\int_{x_i}^{x_{i+1}}f_n|}$$

%%%

My main worry about this proof attempt is that it only shows that a rectangle with height equal to ##f(s_i)## and base ##\Delta x_i## has an area that is within ##\epsilon## units of the area under the graph of ##f## over the region ##[x_i,x_{i+1}]##. But I'm not sure how to expand the proof, if it is correct, in order to show that the area of the entire Riemann sum converges to the integral of ##f##.

Ideally, I would start by building the partition ##P## one point at a time so that the inequality above will hold. But the problem is that I wouldn't know beforehand how many partitions of ##[x_0,x_m]## would be needed. Maybe I could have each partition be of the same length sufficient in order to ensure convergence, then calculate it that way. But I'm a bit hesitant about implementing this idea; what if there is no uniform partition length that will ensure that the area of the rectangle converges to the area under ##f## over that partition?

$$|\sup\{|f-f_n|(x):x\in D\}|<\frac{\epsilon}{3}$$

where ##D## denotes the intersection of the domains of ##f## and ##f_n##.

Choose a partition ##P:=\{x_0,\ldots,x_m\}## with ##x_i<x_{i+1}## where ##0\leq i\leq m-1## s.t. the following holds.

%%% EDIT: The function to be integrated has been corrected from ##f## to ##f_n##.

$$|f_n(s_i)-f_n(t_i)|\Delta x_i<\frac{\epsilon}{3}$$

$$\Delta x_i\leq 1$$

$$|f_n(t_i)\Delta x_i -\int_{x_i}^{x_{i+1}} f_n|<\frac{\epsilon}{3}$$

where ##s_i,t_i\in[x_i,x_{i+1}]## and ##\Delta x_i:=x_{i+1}-x_i##.

Hence,

$$\eqalign{%

\epsilon>&|f_n(s_i)-f(t_i)|\Delta x_i\cr

+&||f-f_n||+|f_n(s_i)\Delta x_i-\int_{x_i}^{x_{i+1}}f_n|\cr

\geq&|f_n(s_i)-f_n(t_i)|\Delta x_i\cr

+&|f(s_i)-f_n(s_i)|+|f_n(t_i)\Delta x_i - \int_{x_i}^{x_{i+1}}f_n|\cr

\geq&|(f_n(s_i)-f_n(t_i))\Delta x_i\cr

+&(f(s_i)+f_n(s_i))\Delta x_i\cr

+&(f_n(t_i)\Delta x_i-\int_{x_i}^{x_{i+1}} f_n)|\cr

=&|f(s_i)\Delta x_i-\int_{x_i}^{x_{i+1}}f_n|}$$

%%%

My main worry about this proof attempt is that it only shows that a rectangle with height equal to ##f(s_i)## and base ##\Delta x_i## has an area that is within ##\epsilon## units of the area under the graph of ##f## over the region ##[x_i,x_{i+1}]##. But I'm not sure how to expand the proof, if it is correct, in order to show that the area of the entire Riemann sum converges to the integral of ##f##.

Ideally, I would start by building the partition ##P## one point at a time so that the inequality above will hold. But the problem is that I wouldn't know beforehand how many partitions of ##[x_0,x_m]## would be needed. Maybe I could have each partition be of the same length sufficient in order to ensure convergence, then calculate it that way. But I'm a bit hesitant about implementing this idea; what if there is no uniform partition length that will ensure that the area of the rectangle converges to the area under ##f## over that partition?

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