Problem: Let $\{f_n\}$ be a sequence of measurable functions that converge pointwise to $f$ on $E$. Let $M \ge 0$ be such that $\int_E f_n \le M$ for all $n$. Show that $\int_E f \le M$. Proof: Since $\int_E f_n \le M$ for all $n$, it must be that $\lim_{n \to \infty} \int_E f_n \le M$. But Fatou's lemma says $\int_E f \le \lim \inf \int_E f_n$. Putting these inequalities together, $\int_E f \le \lim \inf \int_E f_n \le \lim_{n \to \infty} f_n \le M$.
For the most part that's explained by the fact that conservatives tend to be quite ignorant and stupid in this day and age, Zee. Perhaps I'll put you on ignore.
People who need other people to survive lean liberal , academics need society to support them so they value social support and that’s the essence of being a liberal
Okay so if the correct negation is $\lnot P = \forall d \in [a,b], \exists x \in [a,b]: ~f(x) > f(d)$ is the general idea to show that this contradicts continuity of $f$ over $[a,b]$?
@0celo7 I have rewritten it in rational form, and I was hoping to factor the denominator, but it turns out I have to factor a quartic. I tried doing synthetic division, but I suspect the roots are complex. The goal was to see if I could do partial fractions and express as a sum of $\int$
To me, extreme value theorem is "obvious" and yet so much in analysis is about proving things we take for granted so I don't want to get too informal and just assert something as true without proof
this is saying, correct me if I'm wrong, that if there is no maximum, then for every single f(d) in our interval (there are infinitely many of them), there is an even bigger value of f(x) somewhere which must also come from our interval technically. This alone feels like a contradiction to me but I'm not sure if this is enough to "show it"
Take a calculus of variations approach. Consider a sequence $(x_n)\subset [a,b]$ such that $f(x_n)\to \sup f$. Get a convergent subsequence. Then use continuity.