[proofplan]
We work directly from the definition of Hausdorff measure via coverings. Given any countable cover of $E$ by sets of diameter at most $\delta$, the Lipschitz condition produces a cover of $f(E)$ by sets of diameter at most $L\delta$. Comparing the resulting pre-measure sums gives $\mathcal{H}^s_{L\delta}(f(E)) \leq L^s \mathcal{H}^s_\delta(E)$. Taking the infimum over all admissible covers and then sending $\delta \to 0$ yields the result.
[/proofplan]
[step:Lift a cover of $E$ to a cover of $f(E)$ using the Lipschitz condition]
Fix $\delta > 0$ and let $\{C_j\}_{j=1}^\infty$ be a countable collection of sets with $E \subset \bigcup_{j=1}^\infty C_j$ and $\operatorname{diam}(C_j) \leq \delta$ for each $j$. Define $D_j := f(C_j \cap E)$ for each $j$. We claim that $\{D_j\}_{j=1}^\infty$ covers $f(E)$ and each $D_j$ has diameter at most $L\delta$.
**Covering property.** Let $y \in f(E)$. Then $y = f(x)$ for some $x \in E$. Since $E \subset \bigcup_j C_j$, there exists an index $j$ with $x \in C_j$, so $x \in C_j \cap E$ and $y = f(x) \in f(C_j \cap E) = D_j$. Hence $f(E) \subset \bigcup_{j=1}^\infty D_j$.
**Diameter control.** For any two points $y_1, y_2 \in D_j$, there exist $x_1, x_2 \in C_j \cap E$ with $y_i = f(x_i)$. Since $f$ is Lipschitz with constant $L$,
\begin{align*}
|y_1 - y_2| = |f(x_1) - f(x_2)| \leq L |x_1 - x_2| \leq L \operatorname{diam}(C_j) \leq L\delta.
\end{align*}
Taking the supremum over all such pairs gives $\operatorname{diam}(D_j) \leq L \operatorname{diam}(C_j) \leq L\delta$.
[guided]
The entire argument hinges on one observation: the Lipschitz condition $|f(x) - f(y)| \leq L|x - y|$ converts diameter bounds on the domain into diameter bounds on the image. If a set $C$ has $\operatorname{diam}(C) \leq \delta$, then its image $f(C)$ has $\operatorname{diam}(f(C)) \leq L\delta$, because for any $y_1 = f(x_1), y_2 = f(x_2) \in f(C)$ we have
\begin{align*}
|y_1 - y_2| = |f(x_1) - f(x_2)| \leq L|x_1 - x_2| \leq L \operatorname{diam}(C).
\end{align*}
Now fix $\delta > 0$ and let $\{C_j\}_{j=1}^\infty$ be any countable cover of $E$ with $\operatorname{diam}(C_j) \leq \delta$ for all $j$. Set $D_j = f(C_j \cap E)$. Then:
- **$\{D_j\}$ covers $f(E)$:** If $y \in f(E)$, write $y = f(x)$ with $x \in E$. Since $x \in \bigcup_j C_j$, there is some index $j$ with $x \in C_j$, so $y = f(x) \in f(C_j \cap E) = D_j$.
- **Each $D_j$ has controlled diameter:** By the observation above, $\operatorname{diam}(D_j) = \operatorname{diam}(f(C_j \cap E)) \leq L \operatorname{diam}(C_j) \leq L\delta$.
Why do we use $C_j \cap E$ rather than $C_j$ in the definition of $D_j$? Because $f$ is only defined on $A \supset E$, and we only need the image of points in $E$. This makes $D_j = f(C_j \cap E) \subset f(E)$, which is harmless since we only need the covering property.
[/guided]
[/step]
[step:Compare the pre-measure sums and pass to the infimum]
Since $\{D_j\}$ covers $f(E)$ and $\operatorname{diam}(D_j) \leq L\delta$, this is an admissible cover for $\mathcal{H}^s_{L\delta}(f(E))$. Therefore
\begin{align*}
\mathcal{H}^s_{L\delta}(f(E)) \leq \sum_{j=1}^\infty \alpha(s) \left(\frac{\operatorname{diam}(D_j)}{2}\right)^s \leq \sum_{j=1}^\infty \alpha(s) \left(\frac{L \operatorname{diam}(C_j)}{2}\right)^s = L^s \sum_{j=1}^\infty \alpha(s) \left(\frac{\operatorname{diam}(C_j)}{2}\right)^s.
\end{align*}
The cover $\{C_j\}$ was arbitrary among all countable covers of $E$ with diameters at most $\delta$. Taking the infimum over all such covers on the right-hand side gives
\begin{align*}
\mathcal{H}^s_{L\delta}(f(E)) \leq L^s \, \mathcal{H}^s_\delta(E).
\end{align*}
[guided]
Recall the definition of the Hausdorff pre-measure: for any set $S \subset \mathbb{R}^n$ and $\eta > 0$,
\begin{align*}
\mathcal{H}^s_\eta(S) = \inf \left\{ \sum_{j=1}^\infty \alpha(s) \left(\frac{\operatorname{diam}(C_j)}{2}\right)^s : S \subset \bigcup_{j=1}^\infty C_j, \; \operatorname{diam}(C_j) \leq \eta \right\}.
\end{align*}
We established that $\{D_j\}$ covers $f(E)$ with $\operatorname{diam}(D_j) \leq L\delta$, so $\{D_j\}$ is an admissible cover in the infimum defining $\mathcal{H}^s_{L\delta}(f(E))$. This gives the first inequality:
\begin{align*}
\mathcal{H}^s_{L\delta}(f(E)) \leq \sum_{j=1}^\infty \alpha(s) \left(\frac{\operatorname{diam}(D_j)}{2}\right)^s.
\end{align*}
Since $\operatorname{diam}(D_j) \leq L \operatorname{diam}(C_j)$ and the function $t \mapsto t^s$ is monotone increasing on $[0,\infty)$ for $s \geq 0$, we bound each summand:
\begin{align*}
\alpha(s) \left(\frac{\operatorname{diam}(D_j)}{2}\right)^s \leq \alpha(s) \left(\frac{L \operatorname{diam}(C_j)}{2}\right)^s = L^s \cdot \alpha(s) \left(\frac{\operatorname{diam}(C_j)}{2}\right)^s.
\end{align*}
Summing and then taking the infimum over all admissible covers $\{C_j\}$ of $E$ with $\operatorname{diam}(C_j) \leq \delta$ yields $\mathcal{H}^s_{L\delta}(f(E)) \leq L^s \mathcal{H}^s_\delta(E)$.
[/guided]
[/step]
[step:Send $\delta \to 0$ to obtain the Hausdorff measure bound]
As $\delta \to 0$, also $L\delta \to 0$ (here we use $L \geq 0$; if $L = 0$ then $f$ is constant and $f(E)$ is a single point, so $\mathcal{H}^s(f(E)) = 0$ and the inequality holds). By definition, $\mathcal{H}^s(S) = \lim_{\eta \to 0^+} \mathcal{H}^s_\eta(S)$ for any $S$. The pre-measures $\mathcal{H}^s_\eta(S)$ are monotone non-decreasing as $\eta \to 0^+$ (smaller $\eta$ restricts the class of admissible covers), so the limit exists. Taking $\delta \to 0$ in the inequality $\mathcal{H}^s_{L\delta}(f(E)) \leq L^s \mathcal{H}^s_\delta(E)$ gives
\begin{align*}
\mathcal{H}^s(f(E)) = \lim_{\delta \to 0^+} \mathcal{H}^s_{L\delta}(f(E)) \leq L^s \lim_{\delta \to 0^+} \mathcal{H}^s_\delta(E) = L^s \, \mathcal{H}^s(E).
\end{align*}
[/step]
