Unknown work of Nöbeling on topological/Hausdorff dimension Let $\mathcal{H}^n$ denote the Hausdorff measure, $\dim_H X$ the Hausdorff dimension, and $\dim X$ the topological dimension of $X$. 
A well known result of
Szpilrajn (He changed his name to Marczewski while hiding from Nazi persecution) proved in [S] asserts that
if $\mathcal{H}^{n+1}(X)=0$, then  the topological dimension of $X$ is at most $n$. 
Szpilrajn's proof is reproduced in [Theorem 7.3, HW] and [Theorem 8.15, H]. 
Szpilrajn however, mentions in [S] that his argument is based on Nöbeling's proof of a weaker result that topological dimension is bounded from above by the Hausdorff dimension of a metric space. However, he did not provide any reference to Nöbeling's work. 
There is also no reference to Nöbeling's work in the book by Hurewicz and Wallman. 

Question. Does anybody know the reference to the original work of Nöbeling?

[H] J. Heinonen,
Lectures on analysis on metric spaces. Universitext. Springer-Verlag, New York, 2001.
[HW] W. Hurewicz, H. Wallman, Dimension Theory. Princeton Mathematical Series, v. 4. Princeton University Press, Princeton, N. J., 1941.
[S] E. Szpilrajn, La dimension et la mesure, Fund. Math. 28 (1937), 81--89.
 A: It seems to be:


*

*Nöbeling, G., Hausdorffsche und mengentheoretische Dimension, Ergebnisse math. Kolloquium Wien 3, 24-25 (1931). ZBL57.0749.02.


Google shows the first and sporadically the second page (19 lines) from this reprint:


*

*Menger, Karl, Results of a mathematical colloquium, Wien: Springer. ix, 470 S. (1998). ZBL0917.01024.

A: 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. $\Box$
