[proofplan]
The graph $\Gamma_f(A) = \{(x, f(x)) : x \in A\}$ is the image of $A$ under the graph map $\Phi: A \to \mathbb{R}^{m+n}$ defined by $\Phi(x) = (x, f(x))$. We show that $\Phi$ is Lipschitz with constant $\sqrt{1 + L^2}$, then apply the [Lipschitz Bound on Hausdorff Measure](/theorems/3056) with exponent $s = m$, and finally invoke the identity [$\mathcal{H}^m = \mathcal{L}^m$ on $\mathbb{R}^m$](/theorems/3048) to convert $\mathcal{H}^m(A)$ to $\mathcal{L}^m(A)$.
[/proofplan]
[step:Define the graph map $\Phi$ and compute its Lipschitz constant]
Define the graph map
\begin{align*}
\Phi: A &\to \mathbb{R}^{m+n} \\
x &\mapsto (x, f(x)).
\end{align*}
The image of $\Phi$ is precisely $\Gamma_f(A)$. For any $x, y \in A$, the Euclidean norm in $\mathbb{R}^{m+n} = \mathbb{R}^m \times \mathbb{R}^n$ gives
\begin{align*}
|\Phi(x) - \Phi(y)|^2 = |x - y|^2 + |f(x) - f(y)|^2.
\end{align*}
Since $f$ is Lipschitz with constant $L$, we have $|f(x) - f(y)| \leq L|x - y|$, so
\begin{align*}
|\Phi(x) - \Phi(y)|^2 \leq |x - y|^2 + L^2|x - y|^2 = (1 + L^2)|x - y|^2.
\end{align*}
Taking square roots: $|\Phi(x) - \Phi(y)| \leq \sqrt{1 + L^2} \, |x - y|$. Hence $\Phi$ is Lipschitz with constant $\sqrt{1 + L^2}$.
[guided]
The graph $\Gamma_f(A)$ is parametrized by $A$ via the map $\Phi(x) = (x, f(x))$. To control the Hausdorff measure of the graph, we want to apply the [Lipschitz Bound on Hausdorff Measure](/theorems/3056) to $\Phi$. For this, we need to know the Lipschitz constant of $\Phi$.
In $\mathbb{R}^{m+n}$, the distance between two graph points $(x, f(x))$ and $(y, f(y))$ is
\begin{align*}
|\Phi(x) - \Phi(y)|^2 = |x - y|^2 + |f(x) - f(y)|^2.
\end{align*}
The identity component $x \mapsto x$ contributes $|x - y|^2$, and the $f$ component contributes at most $L^2|x - y|^2$. The two contributions are non-negative, so
\begin{align*}
|\Phi(x) - \Phi(y)|^2 \leq (1 + L^2)|x - y|^2.
\end{align*}
This gives $\operatorname{Lip}(\Phi) \leq \sqrt{1 + L^2}$.
Note also that $|\Phi(x) - \Phi(y)| \geq |x - y|$ (since the first component alone gives $|x-y|$), so $\Phi$ is injective. This is not needed for the upper bound but will matter when we discuss bi-Lipschitz maps and the area formula.
[/guided]
[/step]
[step:Apply the Lipschitz bound and convert $\mathcal{H}^m$ to $\mathcal{L}^m$]
By the [Lipschitz Bound on Hausdorff Measure](/theorems/3056) applied to $\Phi: A \to \mathbb{R}^{m+n}$ with Lipschitz constant $\sqrt{1+L^2}$ and exponent $s = m$:
\begin{align*}
\mathcal{H}^m(\Gamma_f(A)) = \mathcal{H}^m(\Phi(A)) \leq (\sqrt{1 + L^2})^m \, \mathcal{H}^m(A) = (1 + L^2)^{m/2} \, \mathcal{H}^m(A).
\end{align*}
Since $A \subset \mathbb{R}^m$ is a Borel set, [$\mathcal{H}^m = \mathcal{L}^m$ on $\mathbb{R}^m$](/theorems/3048) gives $\mathcal{H}^m(A) = \mathcal{L}^m(A)$. Substituting:
\begin{align*}
\mathcal{H}^m(\Gamma_f(A)) \leq (1 + L^2)^{m/2} \, \mathcal{L}^m(A).
\end{align*}
[guided]
We apply the [Lipschitz Bound on Hausdorff Measure](/theorems/3056), which states: if $g: B \to \mathbb{R}^k$ is Lipschitz with constant $M$ and $s \geq 0$, then $\mathcal{H}^s(g(E)) \leq M^s \mathcal{H}^s(E)$ for every $E \subset B$. We apply this with $g = \Phi$, $B = A$, $k = m + n$, $M = \sqrt{1 + L^2}$, $s = m$, and $E = A$. The hypotheses are satisfied: $\Phi$ is Lipschitz (verified in the previous step), and $A \subset \mathbb{R}^m$ is a subset. This gives
\begin{align*}
\mathcal{H}^m(\Phi(A)) \leq (\sqrt{1 + L^2})^m \, \mathcal{H}^m(A).
\end{align*}
The final step is the identification $\mathcal{H}^m(A) = \mathcal{L}^m(A)$. This requires the theorem [$\mathcal{H}^m = \mathcal{L}^m$ on $\mathbb{R}^m$](/theorems/3048), which identifies the $m$-dimensional Hausdorff measure on $\mathbb{R}^m$ with the Lebesgue measure. The proof of that identity relies on the [Isodiametric Inequality](/theorems/3047) — the fact that among all sets of a given diameter in $\mathbb{R}^m$, the ball maximizes Lebesgue measure. Since $A$ is a Borel set in $\mathbb{R}^m$, the identity applies and we obtain $\mathcal{H}^m(\Gamma_f(A)) \leq (1+L^2)^{m/2} \mathcal{L}^m(A)$.
[/guided]
[/step]
