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Mar
12
comment what can you conclude about the number of solutions of the linear system Ax = b?
Why would $A$ have a right inverse? It could be that $a_{1},...,a_{4}$ span something lower dimensional that $\mathbb{R}^{3}$. In fact $A$ could be anything from the null matrix to full rank matrix here.
Mar
12
comment If $\| f \|_p \leq 1$, then $|f(x)| \leq 1$ for some $x$
Yeah, that would do the job.
Mar
12
comment what can you conclude about the number of solutions of the linear system Ax = b?
@sovon: Which part did you not understand?
Mar
12
comment If $\| f \|_p \leq 1$, then $|f(x)| \leq 1$ for some $x$
@kobe: If $f>g$ then in general $\int f\geq \int g$. The pointwise consideration is not taking in to account any properties of the measure and the integration depends on the measure. I guess his question is how do you conclude here that the inequality is strict with the integrals? Might be trivial in the case of Lebesgue measure, but what is the argument?
Mar
12
revised what can you conclude about the number of solutions of the linear system Ax = b?
added 98 characters in body
Mar
12
comment what can you conclude about the number of solutions of the linear system Ax = b?
@YourAdHere: If you assume that $a_{1},...,a_{4}$ are the columns of $A$, then yes. But in that case there are infinitely many solutions as I pointed out.
Mar
12
answered what can you conclude about the number of solutions of the linear system Ax = b?
Mar
12
comment what can you conclude about the number of solutions of the linear system Ax = b?
What do we know about $a_{1},...,a_{4}$? Are those the columns of $A$?
Mar
10
comment A doubt in Probability essentials by Protter
@user3503589: You're welcome.
Mar
10
comment A doubt in Probability essentials by Protter
@jdods: $X$ is an indicator function in part $(v)$ and this seems to be a discontinuation of it, and rather a general statement. In either case it makes no difference, if one leaves the absolute values the argument goes through with same steps.
Mar
10
comment A doubt in Probability essentials by Protter
@user3503589: There is probably a mistake in the print. What is certain is, that once you leave the absolute values you have an equality, and he is, essentially, arguing that if you drop the absolute values the estimate grows. Which seems to be false. But you can keep the absolute values and the argument goes through.
Mar
10
answered A doubt in Probability essentials by Protter
Mar
10
comment A doubt in Probability essentials by Protter
What is the definition of $p_{\omega}$? Is $\Omega$ a countable space here?
Mar
9
comment Measure of $E$ is the limit of the measure of the open set $\mathcal{O}_n$
@dragon: yeah, precisely what I meant. I'm glad you got it.
Mar
9
revised Measure of $E$ is the limit of the measure of the open set $\mathcal{O}_n$
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Mar
9
comment Measure of $E$ is the limit of the measure of the open set $\mathcal{O}_n$
@dragon: There is no reverse inequality needed to prove. Since $E=\bigcap_{n=1}^{\infty}O_{n}$ and $O_{n}\supseteq O_{n+1}$ for all $n$, then $m(E)=\lim_{n\to\infty}m(O_{n})$ by convergence of measure since $m(O_{1})<\infty$ as $E$ was compact. Are you aware of the result that if $O_{n}\supseteq O_{n+1}$ for all $n$ and $E=\bigcap O_{n}$, then $m(E)=\lim_{n\to\infty}m(O_{n})$ if $m(O_{1})<\infty$?
Mar
9
revised Measure of $E$ is the limit of the measure of the open set $\mathcal{O}_n$
added 483 characters in body
Mar
9
answered Measure of $E$ is the limit of the measure of the open set $\mathcal{O}_n$
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