Pointwise convergent and total variation

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)

Solution 1:

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 $\sup_{i\in I}c_i\leqslant M$).

Solution 2:

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?