Doubts regarding the upper bound for Total Variation I was studying a chapter on Total Variation & Compactness, where I had gone through the following portion:
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We can also relate the total variation with the shifted $L^{1}$-norm. Define:
$\lambda (u,\epsilon) := \int |u(x+\epsilon ) - u(x)| dx$ .
Now, we have the following useful lemma: 
Lemma: Let $u \in L^{1}$. If $\frac{\lambda (u,\epsilon)}{|\epsilon|}$ is bounded as a function of $\epsilon$ , then $u \in B.V.$ & $T.V. (u) = lim_{\epsilon \to 0}\frac{\lambda (u,\epsilon)}{|\epsilon|} $ . Conversely, if $u \in B.V.$, then $\frac{\lambda (u,\epsilon)}{|\epsilon|}$ is bounded as a function of $\epsilon$ & thus $T.V. (u) = lim_{\epsilon \to 0}\frac{\lambda (u,\epsilon)}{|\epsilon|} $ holds.
Proof : Assume first that: $u$ is a smooth function. Let {$x_{i}$} be a partition of the interval in question. Then, 
$|u(x_{i}) - u(x_{i-1})| = |\int_{x_{i-1}}^{x_{i}} u'(x) dx| \leq lim_{\epsilon \to 0}\int_{x_{i-1}}^{x_{i}}|\frac{u(x+\epsilon)-u(x)}{\epsilon}|dx$ .
Summing this over $i$ , we get: $T.V. (u) \leq liminf_{\epsilon \to 0}\frac{\lambda (u,\epsilon)}{|\epsilon|}$ for differentiable functions $u(x)$. Let $u$ be an arbitrary bounded function in $L^{1}$ & $u_{k}$ be a sequence of smooth functions such that $u_{k}(x) \to u(x)$ a.e. & $||u_{k}-u||_{L^{1}} \to 0$ . The triangle inequality shows that:
$|\lambda(u_{k}, \epsilon)-\lambda(u, \epsilon)|\leq 2||u_{k}-u||_{L^{1}} \to 0$ .
[UPTO NOW, I HAVE UNDERSTOOD CLEARLY!!]
Let {$x_{i}$} be a partition of the interval. We can now choose $u_{k}$ such that: $u_{k}(x_{i})=u(x_{i}), \forall i$.
Then, $\Sigma|u(x_{i}) - u(x_{i-1})| \leq liminf_{\epsilon \to 0}\frac{\lambda (u_{k},\epsilon)}{|\epsilon|} $ .
Therefore, $T.V. (u) \leq liminf_{\epsilon \to 0}\frac{\lambda (u,\epsilon)}{|\epsilon|}  $
.....
(& after this there are somethings which are easy PROVIDED I understood a few steps clearly)
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My questions are:


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*How can we choose $u_{k}$ such that: $u_{k}(x_{i})=u(x_{i}), \forall i$ ??

*How does the inequality : $\Sigma|u(x_{i}) - u(x_{i-1})| \leq liminf_{\epsilon \to 0}\frac{\lambda (u_{k},\epsilon)}{|\epsilon|} $ IMPLY the inequality :$T.V. (u) \leq liminf_{\epsilon \to 0}\frac{\lambda (u,\epsilon)}{|\epsilon|}  $ .??


[THE LEFT HAND SIDE IS OKAY I SUPPOSE, BY TAKING SUPREMUM & USING THE DEFINITION OF $T.V.$ , BUT WHAT ABOUT THE R.H.S.??]
Thanking you,
 A: Question 1:  Fix a partition $(x_i)_{i=1,\ldots,n}$ and let $(u_k)_{k \in \mathbb{N}}$ be a sequence of smooth functions such that $u_k \to u$ almost everywhere and $\|u_k-u\|_{L^1} \to 0$.
For fixed $k \in \mathbb{N}$ we set
$$\Delta_i := u(x_i)-u_k(x_i)$$
for $i \in \{1,\ldots,n\}$. Moreover, we pick a smooth function $\chi_k$ such that $0 \leq \chi_k \leq 1$, $\chi_k(0)=1$, $\chi_k(x)=0$ for all $|x| \geq 1/k$. The function
$$\tilde{u}_k(x) := u_k + \sum_{i=1}^n \Delta_i \chi_k(x-x_i)$$
is smooth and satisfies $\tilde{u}_k(x_i) = u(x_i)$ for $i=1,\ldots,n$. Moreover, as $\|\chi_k\|_{L^1} \to 0$ as $k \to \infty$ and $\chi_k(x) \to 0$ for all $x \neq 0$, it follows easily that $\tilde{u}_k(x) \to u(x)$ and $\|\tilde{u}_k-u\|_{L^1} \to 0$ are still satisfied.
Question 2: I'm not even convinced that the inequality
$$\text{T.V.}(u) \leq \lim_{\epsilon \to 0} \frac{\lambda(u,\epsilon)}{\epsilon}$$
holds true; at least, if the standard definition of total variation is used (i.e. the supremum over all partitions). Just consider e.g. $$u(x) := \begin{cases} 1 & x = \frac{1}{2} \\ 0,  & x \neq \frac{1}{2} \end{cases}$$ Then $\text{T.V.}(u)=2$, but $\lambda(u,\epsilon)=0$ for all $\epsilon>0$ since $u=0$ almost everywhere. Actually, it is even worse: The function $u(x) = 1_{\mathbb{Q}}(x)$ shows that the boundedness of $\frac{\lambda(u,\epsilon)}{\epsilon}$ does, in general, not imply that $u$ is of bounded variation.
