Wolfy, the two statements are correct, but your statement is stronger since it applies also to the case $T_F = \infty $. If $T_F$ is finite, then you can proceed as @Qiyu indicates. Qiyu is assuming implicitly that $T_F$ is finite at the statemnt: "Chose some partion $P = \{x_0,\dots,x_n\}$ such that $T-T(P) < \epsilon$ ...", which cannot be done if $T_F = + \infty $.
Your attempt is in the right track, but there is at lest one typo and it needs some cleaning.
Qiyu's proof can be modified to avoid assuming $T_F < \infty $. Here is how.
Choose a partition $P = \{x_0,\dots,x_n\}$ of the interval $[a,b]$ with $a=x_0 < ... < x_n=b$ and choose $\epsilon > 0$.
By the assumption of pointwise convergence, there is $k_0$, that depends on $\epsilon$ and the partition, such that for all $k\ge k_0$ we have $|F_k(x_i) - F(x_i)| < \frac{\epsilon}{2n}$ for $i=0,...,n$.
Then for any $k\ge k_0$, and for $i=1,..,n$
$$|F(x_i) - F(x_{i-1})| \le |F(x_i) - F_k(x_{i})| +
|F_k(x_i) - F_k(x_{i-1})| +
|F(x_{i-1}) - F_k(x_{i-1})| $$
$$ < \frac{\epsilon}{n} + |F_k(x_i) - F_k(x_{i-1})| .$$
Therefore
$$\sum_{1}^{n}|F(x_i) - F(x_{i-1})| < \epsilon +
\sum_{1}^{n}|F_k(x_i) - F_k(x_{i-1})| \le \epsilon + T_{F_k}.$$
Now we can take lim inf over $k \ge k_0$ to get:
$$\sum_{1}^{n}|F(x_i) - F(x_{i-1})| \le \epsilon + \liminf_{j\to \infty}T_j$$
and whether the lim-inf is finite or not, since this holds for any $\epsilon > 0$ we get:
$$\sum_{1}^{n}|F(x_i) - F(x_{i-1})| \le \liminf_{j\to \infty}T_j .$$
Since the last inequality holds for any partition $P$, taking supremum over all partitions we get the result.$\square$
$$ \dots \dots \dots $$
In particular, this shows that point-wise convergence to $F$ + finite lim-inf implies the limit $F$ is of bounded variation.
Let's work a couple of examples. Let's take one of the standard examples of unbounded variations
$$
f(x) = \begin{cases}
\frac{1}{x}\cos(\frac{\pi}{x}) & \text{ for $ x \in (0,1]$},\\
0 & \text{for $x=0$}.
\end{cases}
$$
The graph of $f$ oscilates between the lines $y=x$ and $y=-x$ as $x \rightarrow 0^+.$ Take the partition $P=\{ 0, \frac{1}{n},\frac{1}{n-1},.., \frac{1}{2}, 1 \}$. Since $f(\frac{1}{k}) = \frac{(-1)^k}{k}$ for $k=1,2,..$ we get
$$|f(\frac{1}{k}) - f(\frac{1}{k-1}) | = \frac{1}{k} + \frac{1}{k-1}$$
and
$$\sum_{1}^{n}|f(\frac{1}{k}) - f(\frac{1}{k-1})| = 2*ln(n)*(1+ o(1)) ,$$
therefore $T_f = \infty .$
Let
$$F_u(x) = \begin{cases}
f(x) & \text{ for $x \in (\frac{2}{2u+1},1]$},\\
0 & \text{otherwise}.
\end{cases}
$$
Since $| F_u(x)-f(x)| \le \frac{2}{2u+1}$ for all $x \in [0,1]$, it follows that $F_j\rightarrow f$ uniformly, and pointwise.
$F_u$ is differentiable, except at $x=\frac{2}{2u+1}$, and the derivative is bounded, hence $F_u \in BV([0,1])$. Consider the partition $P=\{ 0, \frac{2}{2u+1}, \frac{1}{u},\frac{1}{u-1},.., \frac{1}{2}, 1 \}$.
Because $F_u(0)=F_u(\frac{2}{2u+1}) = 0$, the variation of $F_u$ over this partition is the same sum as above, i.e.: $$\sum_{1}^{u} \frac{1}{k} +\frac{1}{k-1} = 2*ln(u)*(1+ o(1)) ,$$
so
$\liminf_{u\to \infty}T_{F_u} \ge \liminf_{u\to \infty}2ln(u) = + \infty$. We have an example where both, $\liminf_{u\to \infty}T_{F_u}$ and $T_f$ are infinite. Your re-statement of the exercise in Folland applies.
$$ \dots \dots \dots $$
Let's next take
$$
f_u(x)=\begin{cases}
f(s)& \text{for $x \in (\frac{2}{2u^2+1},\frac{2}{2u+1})$},\\
0 & \text{otherwise}.
\end{cases}
$$
Since $|f_u(x)| \le \frac{2}{2u+1}$, we have $f_u\rightarrow 0$ uniformly, and in hence pointwise as well. Clearly $T_0 = 0$, so the limit function is of bounded variation. Consider the partition with partion points
$$ 0, \frac{2}{2u^2+1}, \text{$\frac{1}{k},$ (for $k=u+1,..,u^2 )$} , \frac{2}{2u+1}, 1 .$$
The variation of $f_u$ over this partition is
$$\sum_{u+1}^{u^2} \frac{1}{k} +\frac{1}{k-1} = 2*ln(u)*(1+ o(1)) ,$$
therefore
$\liminf_{u\to \infty}T_{f_u} \ge \liminf_{u\to \infty}2ln(u) = + \infty$.
This is an example where $\liminf_{u\to \infty}T_{f_u} = + \infty $ and $T_f$ is finite.
