0

enter image description here

Why in the theorem taken a Cauchy in measure sequence and not any sequence that converges in measure? Before state the theorem, the text says "we shall now prove a result due to F. Riesz that implies that if a sequence $(f_n)$ converges in measure to $f$, then some subsequence converges almost everywhere to $f$". Every convergent sequence in measure is cauchy in measure? Otherwise, I do not see how this theorem justifies the fact that in reality for any sequence of functions convergent in measure there is a subsequence that converges almost everywhere.

I saw that to prove this statement it is not even necessary for the sequence to be Cauchy in measure, as shown here.

I would be very grateful if you clarify my doubt.

JadMON2k1
  • 376
  • 1
    It shows that a sequence which is Cauchy in measure actually converges in measure, which is a stronger assertion than starting with a sequence that is convergent in measure. – copper.hat Oct 25 '22 at 04:47
  • @copper.hat Why is a stronger assertion? – JadMON2k1 Oct 25 '22 at 23:37
  • Well, I should have written "apparent" or similar, since being Cauchy (as above) and converging are equivalent. But a priori it is not obvious (to me) that being Cauchy in measure means that there is some point that the sequence converges to (in measure). – copper.hat Oct 25 '22 at 23:40

1 Answers1

1

If $(f_n:n\in\mathbb{N})$ is a sequence in $(X,\mathcal{M},\mu)$ and if it converges in measure to $f$, then $(f_n:n\in\mathbb{N})$ is a Cauchy sequence with respect to convergence in measure $\mu$. Indeed, since $f_n\rightarrow f$ in measure, for every $\epsilon>0$ there exists $N_{\epsilon}>0$ such that $\mu\{x:|f_n(x)-f(x)|\geq\frac{\epsilon}{2}\}<\frac{\epsilon}{2}$ for $n\geq N_{\epsilon}$. Then for $m,n\geq N_{\epsilon}$ we have $$\mu\{x:|f_m(x)-f_n(x)|\geq\epsilon\}\leq\mu\{x:|f_m(x)-f(x)|\geq\frac{\epsilon}{2}\}+ \mu\{x:|f_n(x)-f(x)|\geq\frac{\epsilon}{2}\}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon. $$

mio
  • 1,048
  • 1
  • 4
  • 14