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Aug
14
asked Literature on interpolation in Hardy spaces
Jul
3
revised Question about a proof, monotone differentiation theorem
edited tags
Jul
2
awarded  Curious
Jun
30
comment If there exists an integrable function that is not zero a.e., then the measure is $\sigma$-finite
You're done? You've shown that the space $\Omega$ is a countable union of measurable sets of finite measure under $\mu$, so the space is $\sigma$-finite under $\mu$.
May
8
comment Question about a proof, monotone differentiation theorem
"LiuGang Thanks. That is how I reasoned. But now for a rigorous proof the author sandwiches the difference quotients of $F$ between two other difference quotients, as in my post. I feel like $G(x-)$ could be different from $G(x)$ for a sequence of points $x \rightarrow x_0$ and make the left-most and right-most quotients give different results.
May
7
comment Question about a proof, monotone differentiation theorem
@LiuGang I'm sorry but I miss how this helps. Could you explain a little more?
May
7
asked Question about a proof, monotone differentiation theorem
Apr
14
awarded  Popular Question
Mar
28
comment If $K \leq H \leq G$, show that $[G:K] = [G:H][H:K]$.
$G = \cup_{a \in A} Ha = \cup_{a \in A, b \in B} (Kb)a = \cup_{a \in A, b \in B} Kba$, where A is a transversal for $H$ in $G$, and $B$ is a transversal for $K$ in $H$. Of course $|A| = [G:H]$ and $|B| = [H:K]$. You just need to show that the sets $Kba$ are distinct.
Mar
5
awarded  Popular Question
Oct
12
awarded  Yearling
Jul
21
accepted Proof details of Theorem 11.1 in Atiyah-Macdonald
Jul
20
comment Proof details of Theorem 11.1 in Atiyah-Macdonald
Thank you very much for these references. I was very confused, as for example Lang seems to give the same proof as AM and there $g(t)$ also appears. I'll go through the proofs in the links you posted in the morning and then accept your answer.
Jul
20
comment Proof details of Theorem 11.1 in Atiyah-Macdonald
There must be something simple that I am missing here. Given sequences $0 \rightarrow K_n \rightarrow M_n \xrightarrow{x_s} M_{n+k_s} \rightarrow L_{n+k_s} \rightarrow 0$ I would have that $L_{n+k_s} = M_{n+k_s}/x_sM_n$ by exactness. But then what do I take for $L_n$ where $n = 0, 1, \ldots, k_s-1$? I understood that $L$ is the cokernel of the map on $M$ given by multiplication by $x_s$.
Jul
19
comment Proof details of Theorem 11.1 in Atiyah-Macdonald
@YACP I am sorry, I do not really understand. Perhaps I am wrong in the assumption that $L_n = M_n$ for $n = 0, 1, \ldots, k_s-1$?
Jul
19
asked Proof details of Theorem 11.1 in Atiyah-Macdonald
Jun
15
accepted Atiyah-Macdonald, Proposition 2.12, uniqueness of the tensor product.
Jun
15
comment Atiyah-Macdonald, Proposition 2.12, uniqueness of the tensor product.
Oh, that was indeed very simple. Thanks.
Jun
15
asked Atiyah-Macdonald, Proposition 2.12, uniqueness of the tensor product.
May
10
awarded  Nice Question