
> 
> Let $f:[0,1] \to \mathbb{R}, G = graph(f)$.
> 
> 
> If $\sum\_{j = 1}^m |f(x\_i)-f(x\_{i-1})|^s \le c$ for all partitions $0 = x\_0< \ldots < x\_m = 1 $ then $H^s(G) < \infty$
> 
> 
> 


What technique can I use to prove this result?


Can it be reduced to the theorem stating that a rectifiable curve $\Gamma$ has $H^1(\Gamma) < \infty$?



Hausdorff outer measure is finite if $\sum_{j = 1}^m |f(x_i)-f(x_{i-1})|^s \le c$
Take $s \ge 1$. We will use $(x+y)^s \le x^s+y^s$ for positive $x,y$.


Let $C=2^{1+s/2}(1+c)$. I claim: $\mathcal H^s(G) \le C$. 


Let $N \in \mathbb N$. Let $\eta > 0$ be so small that
$N2^s\eta^s \le c$.


For $j=1,2,\dots$ let
$$
M\_j = \sup\left\{f(x): \frac{j-1}{N} \le x \le \frac{j}{N}\right\},\qquad
m\_j = \inf\left\{f(x): \frac{j-1}{N} \le x \le \frac{j}{N}\right\} .
$$


Now construct a partition $\mathcal P$ of $[0,1]$. First put $0,1 \in \cal P$. Then, for each $j$ from $1$ to $N$, put $x\_j, y\_j \in \cal P$, 
chosen so that $|f(x\_j) - M\_j| < \eta$ and
$|f(y\_j) - m\_j| < \eta$. Then 
$|M\_j - m\_j| \le |f(x\_j)-f(y\_j)| + 2 \eta$. This partition has $2N+2$ dividing points in it. Estimate
\begin{align}
\sum\_{j=1}^N |M\_j-m\_j|^s &\le \sum\_{j=1}^N\big(|f(x\_j)-f(y\_j)|+2 \eta\big)^s
\\ &\le N2^s\eta^s+\sum\_{j=1}^N |f(x\_j))-f(y\_j)|^s \le
N2^s\eta^s+c \le 2c .
\end{align}


Now construct a cover $\mathcal Q$ of $G$. Subdivide the plane into squares of side $1/N$ using vertical lines $x=k/N, k \in \mathbb Z$ and horizontal lines $y=k/N, k \in \mathbb Z$. Let $\mathcal Q$ include those squares that meet
the graph $G$. Let us say an index $j$ is **normal** if $M\_j - m\_j < 1/N$, and **abnormal** if $M\_j - m\_m \ge 1/N$


For a given $j$, how many squares in $\mathcal Q$ are there with base $[(j-1)/N,j/N]$? If $j$ is normal, then there are at most $2$. If $j$ is abnormal
then there are at most
$$
\lceil NM\_j\rceil - \lfloor Nm\_j\rfloor \le N(M\_j-m\_j) +2 .
$$
Now compute
$$
\sum\_{U \in \mathcal Q} \text{diam}(U)^s \le
\left(\frac{\sqrt2}{N}\right)^s \left[\sum\_{j\text{ normal}} 2+
\sum\_{j\text{ abnormal}}\big( N(M\_j-m\_j) +2\big)\right]
$$
But
$\sum\_{j=1}^N 2 = 2N$ and
$$
\sum\_{j\text{ abnormal}} N(M\_j-m\_j) 
\le N \sum\_{j\text{ abnormal}} (M\_j-m\_j)^s N^{s-1}
\le N^s \sum\_{j=1}^N (M\_j - m\_j)^s \le 2cN^s .
$$
Therefore
$$
\sum\_{U \in \mathcal Q} \text{diam}(U)^s \le 
\frac{2^{s/2}}{N^s}\big[2N + 2cN^s\big]
=\frac{2^{1+s/2}}{N^{s-1}} + 2^{1+s/2} c \le 2^{1+s/2} (1+c) = C
$$


Hence $\mathcal H^s\_{\sqrt{2}/N} \le C$. 


This is true for all (large enough) $N$, so $\mathcal H^s(G) \le C$.


