An inequality about bounded variation function Let $T^{b}_{a}\left(f\right)$ stand for total variation of $f$ on the interval $[a,b]$.Suppose that $\left\{f_{n}\right\}$ converges to $f(x)$ pointwise.
Prove that$$T^{b}_{a}\left(f\right)\leq\liminf T^{b}_{a}\left(f_{n}\right)$$
The only tool which I know to prove this is the definition of total bounded variation of a function.And I don't know how to make connection between desired result and the definition.
 A: Let us first recall the definition of total variation. The total variation of a function $f$ over the interval $[a,b]$, denoted $T_{a}^{b}(f)$, is defined as follows:
Suppose $P=\{a=x_{0},\ldots,x_{m}=b\}$ denotes a partition of $[a,b]$. Then,
\begin{eqnarray}
T_{a}^{b}(f)=\sup\limits_{P} \left(\sum\limits_{i=1}^{m}\lvert f(x_{i})-f(x_{i-1})\rvert\right),
\end{eqnarray}
where the supremum is computed across all partitions $P$ such as the one defined above. Also, we need the following fact regarding supremum:
\begin{eqnarray}
\sup\limits_{a\in A,~b\in B}(a+b)\leq \sup\limits_{a\in A}a+\sup\limits_{b\in B}b,
\end{eqnarray}
where $A$ and $B$ are some arbitrary sets. We now proceed to prove the desired result.
We have
\begin{align}
T_{a}^{b}(f)&=\sup\limits_{P} \left(\sum\limits_{i=1}^{m}\lvert f(x_{i})-f(x_{i-1})\rvert\right)\\
&=\sup\limits_{P} \left(\sum\limits_{i=1}^{m}\lvert f(x_{i})-f_{n}(x_{i})+f_{n}(x_{i})-f_{n}(x_{i-1})+f_{n}(x_{i-1})-f(x_{i-1})\rvert\right)\\
&\leq \sup\limits_{P} \left(\sum\limits_{i=1}^{m}\lvert f(x_{i})-f_{n}(x_{i})\rvert\right)+\underbrace{\sup\limits_{P} \left(\sum\limits_{i=1}^{m}\lvert f_{n}(x_{i})-f_{n}(x_{i-1})\rvert\right)}_{T_{a}^{b}(f_{n})}+\sup\limits_{P} \left(\sum\limits_{i=1}^{m}\lvert f_{n}(x_{i-1})-f_{}(x_{i-1})\rvert\right)\\
&=\sup\limits_{P} \left(\sum\limits_{i=1}^{m}\lvert f(x_{i})-f_{n}(x_{i})\rvert\right)+T_{a}^{b}(f_{n})+\sup\limits_{P}\left(\sum\limits_{i=1}^{m}\lvert f_{n}(x_{i-1})-f_{}(x_{i-1})\rvert\right).
\end{align}
Since $f_{n}\rightarrow f$ pointwise, for any partition $P$ and for each $i\in \{0,\ldots,m\}$, we have $\lim\limits_{n\rightarrow \infty} \lvert f(x_{i})-f_{n}(x_{i})\rvert=0$, and thus, the first and last terms on the RHS of the last line of the above equation go to $0$ as $n\rightarrow \infty$. We then have the desired result.
In fact, something stronger is true. If we interchange the roles of $f_{n}$ and $f$ in the above set of equations, we arrive at
\begin{eqnarray}
\lim\sup\limits_{n\rightarrow \infty} T_{a}^{b}(f_{n})\leq T_{a}^{b}(f).
\end{eqnarray}
Combining this and the result sought after in this question, we get $\lim\limits_{n\rightarrow \infty} T_{a}^{b}(f_{n})=T_{a}^{b}(f)$.
