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I'm preparing for a test for real analysis and I came across this problem in Royden's book:

Let $\{f_n\}$ be a sequence of real valued functions on $[a,b]$ that converges pointwisely on $[a,b]$ to the real valued function $f$. Show that $TV(f) \leq \liminf ~TV(f_n)?$

This looks quite similar in form to Fatou's Lemma to me, but can't find any way to establish TV with integration, can anybody please help?

(TV is short for total variation)

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up vote 4 down vote accepted

Fix $a=t_0<t_1<\ldots<t_N=b$ a subdivision of $[a,b]$. We have $$\sum_{j=1}^N\left|f(t_j)-f(t_{j-1})\right |=\lim_{n\to +\infty}\sum_{j=1}^N\left|f_n(t_j)-f_n(t_{j-1})\right|,$$ and let $u_n:=\sum_{j=1}^N\left|f_n(t_j)-f_n(t_{j-1})\right|$, $v_n:=\mathrm{ TV} (f_n)$. Since $u_n\leqslant v_n$ for each $n$, we have $\liminf_{n\to+\infty}u_n=\lim_{n\to+\infty}u_n\leqslant \liminf_{n\to \infty}v_n$. As a consequence, we have $$\sum_{j=1}^N|f(t_j)-f(t_{j-1})| \leqslant \liminf_{n\to \infty} \mathrm{ TV} (f_n).$$ Since the considered subdivision is arbitrary, it follows that $$\mathrm{ TV} (f)\leqslant \liminf_{n\to \infty} \mathrm{ TV} (f_n)$$ (because if $I$ is a set and $c_i\leqslant M$ for each $i$, then $\sup_{i\in I}c_i\leqslant M$).

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Davide and @Nana: Thanks so much, I see how it works now. You guys are awesome! – Vokram Apr 1 '12 at 4:07
    
@Topoguy I have edited and added some details. – Davide Giraudo Jul 22 '15 at 7:52
    
@DavideGiraudo Thanks. It makes sense for me. – Topoguy Jul 22 '15 at 15:44
    
When do you think the equality is true? – Topoguy Jul 22 '15 at 22:46

Hint:

Let $a=x_0\lt \ldots \lt x_n=b$ be a subdivision of $[a,b]$. Let $\varepsilon \gt 0$. Then there is an $M$ such that $$ |f_n(x_k) - f(x_k)|\lt \varepsilon /2,\qquad |f_n(x_{k-1}) - f(x_{k-1})| \lt \varepsilon /2,$$ whenever $M<n$. Then consider

$$\sum_{k=1}^N |f(x_k) - f(x_{k-1})|\leq \sum_{k=1}^N |f(x_k) - f_n(x_{k-1})| +\sum_{k=1}^N |f(x_{k-1}) - f_n(x_{k-1})| \\+\sum_{k=1}^N |f_n(x_k) - f_n(x_{k-1})|.$$

Can you continue?

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