Uniform convergence and lengths Consider the following sequence of piece-wise smooth functions $f_n:[0,2]\rightarrow \mathbb{R}$ (below drawn first two functions by their graphs, and is contunued then in natural way.)
 
The function $f:[0,2]\rightarrow \mathbb{R}$, $f(x)=0$ for all $x$ is the limit of this sequence. 
Q. 1 Note that the lengths of graph of each $f_n$ is $2\sqrt{2}$, but the length of limiting function is not $2\sqrt{2}$; why this happens? (Such problem appears as a puzzle shown below picture in some book, but I was looking it through a sequence of functions, it convergence etc.)
Here sequence $f_n$ converges to $f$ uniformly; but still the length of their graphs do not converges to length of graph of limit of $f_n$. This raises following question:
Q If $f_n:[a,b]\rightarrow \mathbb{R}$ is a sequence of piecewise smmoth functions, and converging uniformly to $f:[a,b]\rightarrow \mathbb{R}$, then under what more conditions on $f_n$, we can guarantee the convergence of lengths of $f_n$ to length of $f$?

 A: Let $\mathscr{F}= \{\,f_\alpha\}_{\alpha \in A}$ be a family of real-valued functions defined on $\left[a,b\right]$. We say that $\mathscr{F}= \{\,f_\alpha\}_{\alpha \in A}$ is absolutely equicontinuous on $\left[a,b\right]$ if and only if for every  $\varepsilon>0$ there exists $\delta>0$ so that
\begin{equation} \sum_{k \mathop = 1}^N \left|\, f_\alpha(b_k)-f_\alpha(a_k)\right|<\varepsilon \; \text{ for all } \alpha \in A \, \text{ whenever }\, \sum_{k \mathop = 1}^N \left(b_k-a_k\right)<\delta ,
\end{equation}
and the intervals $\left(a_k,b_k \right)$, $\, k=1, \ldots, N$ are disjoint subsets of $\left[a,b\right]$.

Let $\{\,f_n\}_{n=1}^\infty$ and $\{g_n\}_{n=1}^\infty$ be the sequences of real-valued functions on $\left[0,2\right]$ such that given $n \in \mathbb{N}$, $\; f_n$ and $g_n$ are respectively defined by
\begin{align}f_n(x) :=
  \begin{cases}
                                x-\frac{2k-2}{n} & \text{ if } x \in \left[\frac{2k-2}{n}_, \,\frac{2k-1}{n}\right],  \\
                                   \frac{2k}{n}-x& \text{ if } x \in \left[\frac{2k-1}{n}_, \,\frac{2k}{n}\right] 
  \end{cases} && (k=1,\ldots,n)
\end{align}
\begin{equation}
\end{equation}
\begin{align}g_n(x) :=
  \begin{cases}
                                   n^{n-1}\left(x-\frac{2k-2}{n}\right)^n & \text{ if } x \in \left[\frac{2k-2}{n}_, \,\frac{2k-1}{n}\right],  \\
                                   n^{n-1}\left(\frac{2k}{n}-x\right)^n & \text{ if } x \in \left[\frac{2k-1}{n}_, \,\frac{2k}{n}\right] 
  \end{cases} && (k=1,\ldots,n)
\end{align}
(i) The family of functions $\mathscr{F}:=\{\,f_n\}_{n \in \mathbb{N}}$ is absolutely equicontinuous on $\left[0,2\right]$.
Let $\varepsilon>0$ be given. Choose $\delta = \varepsilon$. Since $\left|\,f_n(d)-f_n(c) \right| \leq (d-c)$ for all $n \in \mathbb{N}$ and $\left(c,d \right) \subseteq \left[0,2\right]$, it follows that $\mathscr{F}$ is an absolutely equicontinuous family on $\left[0,2\right]$.
(ii) The family of functions $\mathscr{G}:=\{g_n\}_{n \in \mathbb{N}}$ is NOT absolutely equicontinuous on $\left[0,2\right]$.
Take $\varepsilon=\frac{1}{2}(1-e^{-1})$, and let $\delta>0$ be given. By the Archimedean Principle we may find a positive integer $N$ so that $N > \frac{1}{\delta} $. 
Now given a positive integer $n \geq N$, we set $a_{n,k}:=\left(\frac{2k-1}{n} - \frac{1}{n^2} \right)$ and $b_{n,k}:=\frac{2k-1}{n}$ $\, \left(k=1, \ldots,n \right)$. So for example we have that
\begin{aligned} &\sum_{k \mathop = 1}^N \left(b_{N,k}-a_{N,k}\right)=\frac{1}{N}<\delta \;\, \text{, and}
\\& \sum_{k \mathop = 1}^N \left|\, g_N(b_{N,k})-g_N(a_{N,k})\right|= 1-\left(1-\frac{1}{N}\right)^N .
\end{aligned}
Since $\lim_{N \to \infty} \left[1-\left(1-\frac{1}{N}\right)^N \right]=1-e^{-1}$, it follows that for sufficiently large $n \geq N$ we shall have
\begin{aligned} &\sum_{k \mathop = 1}^n \left(b_{n,k}-a_{n,k}\right)=\frac{1}{n}<\delta \;\, \text{, BUT}
\\& \sum_{k \mathop = 1}^n \left|\, g_n(b_{n,k})-g_n(a_{n,k})\right|> \varepsilon .
\end{aligned}
So $\mathscr{G}$ is not an absolutely equicontinuous family on $[0,2]$.
Note: $\mathscr{G}$ is a uniformly equicontinuous family on $[0,2]$ because given $n \in \mathbb{N}$ we have that $|g_n(x)|\leq \frac{1}{n} = h_n(x)$ for all $x \in [0,2]$. To be clear, this implies that $\{g_n\}_{n=1}^\infty$ converges uniformly to $g \equiv 0$ on $[0,2]$, since the sequence of dominating functions $\{h_n\}_{n=1}^\infty$  converges uniformly to $0$. Let $\varepsilon>0$ be given. Since $g_n \to 0$ uniformly on $[0,2]$, we know there is a positive integer $N$ so that $|g_n(x)|<\frac{\varepsilon}{2}$ for all $x \in [0,2]$ whenever $n \geq N$. Hence $|g_n(x)-g_n(y)| \leq |g_n(x)|+|g_n(y)|<\varepsilon$ whenever $n \geq N$, and $x,y \in [0,2]$. So we have the $N-1$ functions $\{g_1, \ldots, g_{N-1} \}$ left to find a uniform delta for that will work for every function in the family $\mathscr G$. We know every function in the finite collection $\{g_1, \ldots, g_{N-1} \}$ is continuous on $[0,2]$, as each of these functions is piecewise continuous with pieces that agree at each endpoint (Pasting/Gluing Lemma). Hence each of these functions is uniformly continuous on $[0,2]$, by: Proof verification for a couple of theorems regarding Lebesgue's number. Thus for each $k=1, \ldots, N-1 \,$ we may find a positive number $\delta_k$ so that $|g_k(x)-g_k(y)|<\varepsilon$ whenever $|x-y|<\delta_k$, and $x,y \in [0,2]$. Setting $\delta = \min \{\delta_1, \ldots, \delta_{N-1} \}$ shows that $\mathscr G$ satisfies the definition of a uniformly equicontinuous family on $[0,2]$.
A: Let $L(\,f)$ denote the length of the graph 
\begin{align}
\Gamma_f = \{t+if(t) \in \mathbb{C} : t \in [0, 2]\} \,.
\end{align}
Let $\{\,f_n\}_{n=1}^\infty$ be a sequence of real-valued continuous functions that converge uniformly on $[0, 2]$ to the function $f:=\lim\limits_{n \to \infty}\,f_n$. 
We have that
$\,L(\,f) = \lim\limits_{n \to \infty} L(\,f_n)$  holds if and only if the sequence of functions $\{\,f_n\}_{n=1}^\infty$ forms on $[0,2]$ an absolutely equicontinuous family such that $f^{'}_n$ converges in measure to $f'$.
I can't give a formal proof of the preceding statement. Though I took the liberty of coding up a worse counterexample sequence $\{g_n\}_{n=1}^\infty$ that I found more insightful
 \begin{align}f_n(x) :=
  \begin{cases}
                                x-\frac{2k-2}{n} & \text{ if } x \in \left[\frac{2k-2}{n}_, \,\frac{2k-1}{n}\right],  \\
                                   \frac{2k}{n}-x& \text{ if } x \in \left[\frac{2k-1}{n}_, \,\frac{2k}{n}\right] 
  \end{cases} && (k=1,\ldots,n)
\end{align}
\begin{equation}
\end{equation}
\begin{align}g_n(x) :=
  \begin{cases}
                                   n^{n-1}\left(x-\frac{2k-2}{n}\right)^n & \text{ if } x \in \left[\frac{2k-2}{n}_, \,\frac{2k-1}{n}\right],  \\
                                   n^{n-1}\left(\frac{2k}{n}-x\right)^n & \text{ if } x \in \left[\frac{2k-1}{n}_, \,\frac{2k}{n}\right] 
  \end{cases} && (k=1,\ldots,n)
\end{align}
\begin{equation} 
\begin{aligned}
& L(g_j) =2j \int_0^{1/j} (1+j^{2j}t^{2j-2})^{1/2} dt \\&
L(g_j) \leq L(g_{j+1})\end{aligned} 
\: \;\: \;(\,j=1,2,\ldots)
\end{equation}
Note: $g_1(x)=f_1(x)$ for all $x \in [0,2].$

I should mention that there is a sequence of functions similar to $\{\,f_n\}_{n=1}^\infty$  $(*)$  described in a comment by ThePortakal, here:
Sequence of functions that converges uniformly in $R$, derivatives does not converge punctually at any point of $R$.

$(*)$ The family of functions described by ThePortakal is absolutely equicontinuous on $[0,2]$, and the graphs corresponding to the functions in this family will have constant length $\pi$ on $[0,2]$.

I have decided to add a picture of the work that went into my original answer (excuse the notation abuse with $\,f_n$ on the right-hand side)


Along with Beginner's interesting question here and ThePortakal's comment in another thread, this Dover book was helpful in motivating the sequence $\{g_n\}_{n=1}^\infty$.

A: Suppose $\{f_n\}_{n=1}^\infty$ is a sequence of real-valued functions such that $\lim_{n \to \infty} f_n(a)$ exists, $f_n'$ converges in measure to $g$ on $\mathbb R$, $|g| \in L^1([a,b]),$ and $\{|f_n'|\}_{n=1}^\infty \subset L^1([a,b])$ forms a uniformly integrable (or rather, $\{f_n\}_{n=1}^\infty$ forms an absolutely equicontinuous) family. Let $f(x):=\lim_{n \to \infty} f_n(a)+\int_a^x g dm$. Let $\varepsilon>0$ be given. Since $\{|g|\} \cup \{|f_n'|\}_{n \in \mathbb N}$ is a uniformly integrable family on $[a,b]$ (when applying the definition of uniformly integrable, we may take $\delta=\min \{\delta_1, \delta_2\}$), there is $\delta>0$ so that
$\int_E |f_n'|dm<\varepsilon$ for all $n \in \mathbb N$ and
$\int_E |g|dm<\varepsilon$ whenever $m(E)<\delta$ and $E \subset [a,b]$. Given $n \in \mathbb N$, let $E_n:=\{x \in [a,b]:|f_n'(x)-g(x)|\geq \varepsilon\}$. Since $f_n' \to g$ in measure, there is $N \in \mathbb N$ so that $m(E_n)<\delta$ whenever $n \geq N.$ So we have $n \geq N$ implies that
\begin{align*}
    \int_{[a,b]} |f_n'-g|dm &\leq \int_{[a,b] \setminus E_n} |f_n'-g|dm + \int_{E_n} |f_n'-g|dm \\&\leq \int_{[a,b] \setminus E_n} |f_n'-g|dm + \int_{E_n} |f_n'|dm +\int_{E_n} |g|dm \\& < (b-a)\varepsilon + \varepsilon + \varepsilon= (b-a+2)\varepsilon.
    \end{align*}
So $\lim_{n \to \infty} \int_{[a,b]} |f_n'-f'|dm =\lim_{n \to \infty} \int_{[a,b]} |f_n'-g|dm =0$, and $f_n$ converges pointwise (actually uniformly* - this observation is just made so we may understandably write $L(f)$ instead of $L(\lim_{n \to \infty} f_n)$) to $f$ on $[a,b]$.
Since
$$
\left| \int_{[a,b]} \sqrt{1+{f_n'}^2} dm - \int_{[a,b]} \sqrt{1+{f'}^2} dm\right| 
\le \int_{[a,b]} |f_n'-f'| dm,
$$
we have that $\lim_{n \to \infty} L(f_n) = L(f)$.
*Suppose $\{f_n\}_{n \in \mathbb N}$ is a uniformly equicontinuous family of real-valued functions on $[a,b]$, $f$ is continuous on $[a,b]$, and $f_n \to f$ pointwise on $[a,b]$ (technically we could dispose of the requirement that $f$ is continuous and only require $\{f_n\}_{n \in \mathbb N}$ to be equicontinuous on $[a,b]$ and still have the same result but the chosen assumptions will already be true in the context above). Let $\varepsilon>0$ be given. We may find $\delta_1>0$ and $\delta_2>0$ so that $|f_n(x)-f_n(y)|<\varepsilon$ for all $n \in \mathbb N$ and $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta=\min\{\delta_1,\delta_2\}$ and $x,y \in [a,b]$. By the Archimedean principle, we may find $M \in \mathbb N$ so that $\frac{b-a}M<\delta$. Given $k \in \{1,\ldots, 2M-1\}$, let $x_k:=a+k\frac{b-a}{2M}$. Given $k \in \{1,\ldots, 2M-1\}$, we may find $N_k \in \mathbb N$ so that $|f_n(x_k)-f(x_k)|<\varepsilon$ whenever $n \geq N_k$ ($f_n \to f$ pointwise). Take $N=\max\{N_1, \ldots, N_{2M-1}\}$. Let $x \in [a,b]$. So $|x-x_k|<\delta$ for some $k \in \{1,\ldots, 2M-1\}$, and we have
$$|f_n(x)-f(x)|\leq |f_n(x)-f_n(x_k)|+|f_n(x_k)-f(x_k)|+|f(x_k)-f(x)|<3\varepsilon$$
whenever $n \geq N$. So $\sup_{x \in [a,b]} |f_n(x)-f(x)| \leq 3\varepsilon$ whenever $n \geq N$. Hence $\lim_{n \to \infty} \sup_{x \in [a,b]} |f_n(x)-f(x)|=0$. In other words, $f_n \to f$ uniformly on $[a,b]$.

Considering the Vitali convergence theorem, these assumptions turn out to be equivalent to saying $f_n' \to f'$ in $L^1([a,b])$** which in some sense seems underwhelming. And since we are dealing with spaces of finite measure, $f_n' \to f'$ uniformly would also imply that $\lim_{n \to \infty} L(f_n) = L(f)$.
**Suppose $\{f_n\}_{n=1}^\infty \subset L^1$, $f \in L^1$, and $\lim_{n \to \infty} \int \left|f_n-f \right|dm =0$. Let $\varepsilon>0$ be given. 
By Corollary $3.6$ (of Folland - what the corollary says should be clear from the rest of this sentence), we may find $\delta'>0$ so that $\left| \int_E f d m\right|< \varepsilon$ whenever $m(E)<\delta'$. Since $\lim_{n \to \infty} \int \left|f_n-f \right|dm =0$, there is $N \in \mathbb N$ so that $\int \left|f_n-f \right|dm<\varepsilon$ whenever $n \geq N$. So if $n \geq N$, then we have
  \begin{align*}
 \left| \int_E f_n d m \right| &\leq   \left| \int_E f_n-f d m\right|+\left| \int_E f d m \right|  \\&\leq \int_E \left|f_n-f \right|dm +\left| \int_E f d m\right| \\&\leq \int \left|f_n-f \right|dm +\left| \int_E f d m\right| \\&<\varepsilon+\varepsilon=2\varepsilon.
  \end{align*}
Given $j \in \{1, \ldots, N-1\}$, we may find $\delta_j>0$ so that $\left| \int_E f_j d m \right|< \varepsilon$ whenever $m(E)<\delta_j$ (Corollary $3.6$). Let  $\delta''=\min\{\delta_1, \ldots, \delta_{N-1}\}$. Thus $\left| \int_E f_j d m\right|< \varepsilon$ for all $j \in \{1, \ldots, N-1\}$ whenever $m(E)<\delta''.$ Let $\delta=\min \{\delta', \delta''\}$.
Thus $\left| \int_E f_n d m\right|< \varepsilon$ for all $n \in \mathbb N$ whenever $m(E)<\delta.$ In other words, $\{f_n\}_{n \in \mathbb N}$ is uniformly integrable.
Let $E_n=\{x: |f_n(x)-f(x)|\geq \varepsilon \}$. Since $\lim_{n \to \infty} \int \left|f_n-f \right|dm =0$, there is $N' \in \mathbb N$ so that $\int \left|f_n-f \right|dm<\varepsilon^2$ whenever $n \geq N'$. So if $n \geq N'$, then we have
$$\varepsilon m(E_n) \leq \int_{E_n} \left|f_n-f \right|dm \leq \int \left|f_n-f \right|dm <\varepsilon^2.$$ In other words, $f_n \to f$ in measure.
