Theorem : Let $ f_n \in R([a,b]) $ for all $ n \in \mathbb{N} $ ( $ \{ f_n \} : N \to ( [a,b] \to \mathbb{R} ) $ is a sequence of functions ), and let $ f:[a,b] \to
\mathbb{R} $ and suppose $f_n(x)^\rightarrow_\rightarrow f(x) $. Then $ f \in R([a,b]) $ and $ \int_a^b{f(x)dx} = \lim_{n \to \infty } \int_a^b f_n(x)dx $.
Proof: The proof is separated into two parts,
Proof of integrability:
in order to prove integrability, we'll prove $ \forall \epsilon > 0 $ $ \exists \delta > 0$ s.t. for every partition $ \prod $ of $ [a,b] $ s.t. $ \lambda(\prod) < \delta $ then $ \omega(f,\prod) < \epsilon $ ( This is Darboux's definition for integrability ).
Let $ \epsilon >0 $ be arbitrary, from uniform convergence there exists $ N \in \mathbb{N} $ s.t. $\forall x \in [a,b] $ and $ \forall n \geq N $ we have $ |f_n(x) – f(x) | < \frac{\epsilon}{4(b-a)} $.
Let $ n \geq N $ be arbitrary, $ f_n $ is Riemann Integrable, thus there exist $ \delta > 0 $ s.t. for every partition $ \prod $ s.t. $ \lambda(\pi) < \delta $ then $ \omega(f_n,\prod) < \epsilon/2 $ ( Usage of Darboux's definition for integrability ) . Let $ \prod $ be an arbitrary partition of $ [a,b] $ and suppose $ \lambda(\prod) < \delta $, thus $ \omega(f_n,\prod) < \epsilon / 2 $
Specifically, for every interval $ J \subseteq [a,b] $ we'll have $ | \omega(f,J) – \omega(f_n,J) | \leq \frac{\epsilon}{2(b-a)} $.
Notice that $ \omega(f,\prod) = \sum \omega(f,[x_{i-1},x_i]) \Delta x_{i} \leq \sum \omega(f_n,[x_{i-1},x_i]) \Delta x_{i} + \epsilon/2 < \epsilon $ , and we're finished.
Proof of $ \int_a^b{f(x)dx} = \lim_{n \to \infty } \int_a^b f_n(x)dx $ :
[ there is proof here but I omitted it because my question's not about it ]
My question:
I didn't understand the line where it says " Specifically, for every interval $ J \subseteq [a,b] $ we'll have $ | \omega(f,J) – \omega(f_n,J) | \leq \frac{\epsilon}{2(b-a)} $. " , How did they reach the last inequality?
Notes about notation:
- $ f \in R([a,b]) $ means $ f$ is Riemann integrable on $[a,b] $ .
- $ \lambda(\prod) = max_{i=1,…,n}|{ \triangle x_i}| $ ( this is the mesh of partition $ \prod $ )
- $\omega(f, J)=\sup _{J} f-\inf _{J} f=\sup _{x, y \in J}(f(x)-f(y))$, where J is an interval.
- $\omega(f, \Pi)=\sum_{i=1}^{n} \omega\left(f,\left[x_{i-1}, x_{i}\right]\right) \Delta x_{i}$ , where $ \Pi $ is a partition of some closed interval.
- $f_n(x)^\rightarrow_\rightarrow f(x) $ means the sequence of functions $ \{ f_n \} $ uniformly converges to $ f $
Thanks in advance for help!
Best Answer
For the given $n \geqslant N$, we have for all $x \in [a,b]$,
$$\tag{1}|f(x) - f_n(x)| < \frac{\epsilon}{4(b-a)},$$
and, thus, for any $x,y \in J \subseteq [a,b]$,
$$\tag{2}f(x) < f_n(x)+ \frac{\epsilon}{4(b-a)},$$ $$\tag{3}f(y) > f_n(y)- \frac{\epsilon}{4(b-a)}$$
Subtracting (2) - (3), we get
$$f(x) - f(y) < f_n(x) - f_n(y) + \frac{\epsilon}{2(b-a)}$$
Applying the triangle inequality, it follows that
$$|f(x) - f(y)| < |f_n(x) - f_n(y)| + \frac{\epsilon}{2(b-a)} \leqslant \sup_{x,y \in J}|f_n(x) - f_n(y)| + \frac{\epsilon}{2(b-a)}$$
Taking the supremum on the LHS, we get
$$\sup_{x,y \in J}|f(x) - f(y)| \leqslant \sup_{x,y \in J}|f_n(x) - f_n(y)| + \frac{\epsilon}{2(b-a)}$$
Using the definition of $\omega(\cdot,J)$ and rearranging we get,
$$\tag{4}\omega(f,J) - \omega(f_n,J) \leqslant \frac{\epsilon}{2(b-a)}$$
It also follows from (1) that $$\tag{5}f_n(x) < f(x)+ \frac{\epsilon}{4(b-a)},$$ $$\tag{6}f_n(y) > f(y)- \frac{\epsilon}{4(b-a)},$$
Repeating the previous argument yields
$$\tag{7}\omega(f_n,J) - \omega(f,J) \leqslant \frac{\epsilon}{2(b-a)}$$
Together, (4) and (7) imply that
$$|\omega(f,J) - \omega(f_n,J)| \leqslant \frac{\epsilon}{2(b-a)}$$