# Uparrow sign (convergence?)

Im not entirely sure what this definition means, whilst I'm reading up.

Let $X_n \in$ some sigma algebra $\mathcal{F}$.

$X_n \uparrow X = X_n \subseteq X_{n+1}, \forall n \in \mathbb{N}$ and $\cup X_n = X.$

All that was written in my notes was the above line (which seems to be conditions?) But does $X_n \uparrow X$ mean that $X_n$ converges to $X$?

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It probably means just what you wrote, i.e. that the sets $X_n$ are increasing and that their union is $X$. – mrf Mar 12 '12 at 22:27
It converges in the sense that $\limsup_n X_n=X=\liminf_n X_n$ where $\limsup_nX_n=\bigcup_{n\geq 0}\bigcap_{k\geq n}X_k$ and $\liminf_nX_n=\bigcap_{n\geq 0}\bigcup_{k\geq n}X_k$. – Davide Giraudo Mar 12 '12 at 22:27
It means that $X_n$ increases and converges to $X$. To be specific, for a sequence of subsets $X_n$ of $X$ we define \begin{align*} \limsup_n X_n & = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty} X_m \\\liminf_n X_n & = \bigcup_{n=1}^{\infty} \bigcap_{m=n}^{\infty} X_m. \end{align*} Now we say $\lim_n X_n = X$ if $\limsup_n X_n = \liminf_n X_n = X$. If $X_n$ is monotone, we can prove that $\lim X_n$ always exists. Thus the notation $X_n \uparrow X$ means that $X_1 \subset X_2 \subset X_3 \subset \cdots$ and $X = \bigcup_{n=1}^{\infty} X_n$. – Sangchul Lee Mar 12 '12 at 22:31
Thanks guys, great help, I understand it now! – Gary Mar 12 '12 at 22:46

The line you quote is a definition of $X_n \uparrow X$ although I agree that this is not very clear. That is, the line states that $X_n \uparrow X$ means $X_n \subseteq X_{n+1}$ and $\bigcup X_n = X$.
You should think of $\uparrow$ as meaning increasing pointwise convergence. More generally, if $f_n$ is a sequence of measurable real-valued functions, we say that $f_n \uparrow f$ if $f_n \le f_{n+1}$ pointwise and $f_n \to f$ pointwise. This reduces to the above definition if we take $f_n = 1_{X_n}, f = 1_X$.