All Comments

Bokuan Li
Yesterday at 03:43
Missing paragraph break before (3) $\Rightarrow$ (1).
Bokuan Li
Yesterday at 03:42
Typos: Let $K_{0} \subset K$ be closed, extreme subset of $K$ [...]. Since $K$ is compact, for any chain [...].
Bokuan Li
June 14th at 21:54
Typo: duplicate ”since” and addition spelling.
Bokuan Li
June 5th at 05:30
If it’s a Bochner integral why does Takesaki justify its existence with a Riemann-Stieltjes sum? I think you can get away by skipping the justification entirely.
Bokuan Li
June 5th at 05:16
Reference to Gelfand Mazur seems broken.
Bokuan Li
May 27th at 05:34
Equation 5.1 seems too long to fit on my screen.
Bokuan Li
May 27th at 05:33
My original phrasing was bad in the ”there exists continuous seminorms”. Since these are precisely the seminorms that bound the product, you should state that in the beginning.
Bokuan Li
May 27th at 05:32
Why are your $||\cdot||$s short? How do you do that?
Bokuan Li
May 27th at 05:30
”For any $P \le Q \in \mathscr{P}([a, b])$, then” doesn’t seem correct? Maybe you can just get away with ”Observe” again.
Bokuan Li
May 27th at 05:26
Typo: should be $\tau(\mathcal{S}) \subset \bigcap_{\tau \supset \mathcal{S}}\tau$.
Bokuan Li
May 27th at 05:24

You probably don’t need $* \in \{\cap, \cup\}$ if you only use it for $\cap$. The union for (O2) shouldn’t be over the product $I_{j} \times J$, since $I_{j}$ depends on $J$. Instead, it should be something like: Let $\{U_{j}\}_{j \in J}\subset \tau(\mathcal{B})$. For each $j \in J$, let $I_{j}$ such that $U_{j} = \bigcup_{i \in I_j}B_{i, j}$, then

\[\bigcup_{j \in J}U_{j} = \bigcup_{j \in J}\bigcup_{i \in I_j}B_{i, j}= \bigcup_{(k, f) \in J \times \prod_{j \in J}I_j}B_{k, f\paren{k}}\]
Bokuan Li
May 27th at 05:18
Typo on (T4). Should be $A \subset U$ instead of $A \in U$.
Bokuan Li
May 27th at 05:17
Typo in the name, should be ”generated”. I don’t think you need the ideal assumption here (I use it on my end because it fits with my system).
Bokuan Li
May 26th at 04:12
Off the top of my head, Folland should also require the additional condition, which would imply (5).
Jerrylicious
May 26th at 03:26
Missing $<$ on point (4).