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Let ${f_{n}}$ a increasing sequence of measurable functions such that $f_{n} \rightarrow f$ in measure. Show that $f_{n}\uparrow f$ almost everywhere

My attempt

The sequence ${f_{n}}$ converges to f in measure if for any $\epsilon >0$ there exists $N\in \mathbb{N}$ such that for all n>N, $$ m (\{x: |f_n(x) - f(x)| > \varepsilon\}) \rightarrow 0\text{ as } n \rightarrow\infty. $$

I think that taking a small enough epsilon are concludes the result since $f_n$ are increasing

Can you help me solve this exercise?

Thanks for your help

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I reformulated your question so that it feels right. Can you agree with me? You wrote before that the functions were increasing (which didn't make sense), I re-wrote it to say that the sequence was increasing, not the functions. –  Patrick Da Silva Jun 15 '13 at 3:42
    
Thanks, I think is better understood –  Miguel Mora Luna Jun 15 '13 at 3:46

2 Answers 2

up vote 1 down vote accepted

If $f_n$ converges to $f$ in measure, we have a subsequence $f_{n_k}$ converging to $f$ almost everywhere. As $f_n$ is increasing, we know that $f_n$ converges at every point or every subsequence increases to $\infty$. But as the subsequence $f_{n_k}$ converges to $f$ at almost every point, we have that $f_n$ converges to $f$ almost everywhere.

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Yeah... I feel like all the darkness is hidden in "$f_n$ converges to $f$ in measure hence $f_{n_k}$ converges to $f$ a.e." but then again I think this result is pretty standard. +1 –  Patrick Da Silva Jun 21 '13 at 3:18

The sequence $f_n$ is increasing, hence for each $x \in X$ (I assume $X$ is your measure space), $\{f_n(x)\}$ is an increasing sequence, thus either converges or goes to $+\infty$.

Let $\varepsilon > 0$. The set of values over which $f_n(x) \underset{n \to \infty}{\longrightarrow} L > f(x) + \varepsilon$ (with $L \in \mathbb R \cup \{+\infty\}$) must have measure zero because the sets $$ A_n^{\varepsilon} = \{ x \in X \, | \, f_n(x) > f(x) + \varepsilon \} $$ are such that $$ m(A_n^{\varepsilon}) \le m( \{ x \in X \, | \, |f_n(x) - f(x)| > \varepsilon \} ) \underset{n \to \infty}{\longrightarrow} 0, $$ hence since $f_n(x) \underset{n \to \infty}{\longrightarrow} L$ implies that there exists an $N$ with $f_N(x) > f(x) + \varepsilon$ and that for this $N$, for all $n \ge N$, $f_n(x) \ge f_N(x) > f(x)+\varepsilon$, we have $$ \{ x \in X \, | \, f_n(x) \to L > f(x) + \varepsilon \} = A^{\varepsilon} \subseteq \bigcup_{N \in \mathbb N} \bigcap_{n \ge N} A_n^{\varepsilon} = \liminf A_n^{\varepsilon}, $$ and by Fatou's lemma, $$ m(A^{\varepsilon}) \le m \left( \liminf A_n^{\varepsilon} \right) \le \liminf m(A_n^{\varepsilon}) = \lim m(A_n^{\varepsilon}) = 0. $$ It follows that $$ A = \{ x \in X \, | \, f_n(x) \to L > f(x) \} = \bigcup_{m \in \mathbb N} A^{1/m} $$ and thus $m(A) \le \sum_{m \in \mathbb N} m(A^{1/m}) = 0$. Repeat similarly with $$ B_n^{\varepsilon} = \{ x \in X \, | \, f_n(x) \to L < f(x) - \varepsilon \} $$ to conclude that $f_n(x) \to f(x)$ almost everywhere. (The direction with $B$ should be easier though, but the ideas are the same ; you might not need Fatou's lemma, but I didn't work out the details. It just feels easier.)

Hope that helps,

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