So, the sought for paper is:

Nöbeling, G., Hausdorffsche und mengentheoretische Dimension, Ergebnisse math. Kolloquium Wien 3, 24-25 (1931).

And here is a ``translation" (to English and to modern math exposition standards, if such a thing exists.) What is called the set-theoretical dimension is defined inductively: dim of the empty set is $-1$, and we set $\dim M = k$ if $k$ is the least integer with the property that every point of $M$ has arbitrarily small (open) neighborhoods whose dim is $k-1$.

## Hausdorff and set-theoretical dimension. By George Nöbeling.

Let $M$ be a subset of the Euclidean $\mathbb{R}^n$. One covers $M$ by finitely or countably (infinitely) many balls $K_j$, with diameters $d_j < \rho$, for $p \leq n$ (any non-negative number) form the sum $ \sum d_j^p $. Let $ L_p(\rho, M)$ be the infimum of such sums for all such coverings. Put
$$
L_p (M) = \lim_{\rho \to 0} L_p (\rho, M) .
$$
Obviously there is exactly one number $p= p (M)$ such that for every $q > p$, $L_q (M) = 0$ and for every number $q <p$, $L_q (M) = \infty$. We call this clearly defined number $p$ the Hausdorff dimension of the set $M$ and claim that

**THEOREM**: For any set $M$, the Hausdorff dimension is at least equal to the set-theoretical dimension.

**PROOF.** The proof is done by induction on the set-theoretical dimension $\dim M$. For every $k \in \{0,1,2,\cdots\} $ we prove that if $M$ is a set with $\dim M \geq k$ then $p(M) \geq k$.

For $k=0$ this is obvious thanks to $ p (M) \geq 0.$ Suppose the claim is true for $k$. We must thus prove that if $M$ is any set with $\dim M \geq (k + 1)$, then $p(M) \geq k+1$.

Since $M$ is at least $(k+1)$-dimensional, there exit a point $P$ of $M$ and a number $r_0$, such that for every $(n-1)$-dimensional sphere $S_r$, with the radius $r \leq r_0$ the intersection $M \cap S_r$ is at least $k$-dimensional. [*Otherwise, every point of $M$ would have arbitrarily small open neighborhoods whose boundaries had dim $k-1$ or less, and so by definition $M$ would have dimension less than or equal to $k$.*], and therefore according to the induction hypothesis that
$$
\forall r \leq r_0, \; p (M \cap S_r) \geq k \, .
$$

For each $i \in \mathbb{N}$, let $\{K_{ij}\}_j$ be a covering of the set $M$ by spheres of diameter $d_{ij} < \frac{1}{i}$. For a number $q < k +1$ and an $0 < x \leq r_0$ we set
$$
f_{ij} (x) =
\begin{cases}
d_{ij}^{q-1} & \text{if $S_x \cap K_{ij} \neq \emptyset $ ,}\\
0 & \text{Otherwise.}\\
\end{cases}
$$

We also set
$$
s_i (x) = \sum f_{ij} (x).
$$
Obviously,
$$
s_i(x) \geq L_{q-1}(1/i,S_x \cap M) \, .
$$
Since $ p (M \cap S_r) \geq k > q-1 $, it follows from the induction hypothesis that
$$
\forall x \in (0,r_0], \; \lim_{i \to \infty} s_i(x) = \infty \, .
$$

Thus,
$$
\sum_j \int_0^{r_0} f_{ij}(x) \, dx = \int_0^{r_0} s_i(x) \, dx \xrightarrow{i \to \infty} \infty \, .
$$
Now observe that $f_{ij} (x) = d_{ij}^{q-1}$ for $x$ from an interval whose length is at most $d_{ij}$ -- the diameter of $K_{ij}$ -- and otherwise $f_{ij} (x) = 0.$ Therefore,
$$
\sum_j d_{ij}^{q} = \sum_j \int_0^{d_{ij}} d_{ij}^{q-1} \, dx \geq \sum_j \int_0^{r_0} f_{ij}(x) \, dx \, .
$$
and hence,
$$
\sum_j d_{ij}^{q} \xrightarrow{i \to \infty} \infty.
$$
Since this is true for any coverings $K_{ij}$, we conclude
$$
L_q(M) = \infty \implies p(M) \geq q \, .
$$

Since $q <k + 1$ was arbitrary we have shown that $p (M) \geq k + 1$. This concludes the induction and proves our theorem. $\Box$