Total boundness of Lipschitz densities In the article Almost Sure Testability of Classes of Densities by Devroye and Lugosi in 1999. They claim in Example 10 (page 9) that Lipschitz densities on [0,1] with Lipschitz constant bounded by some $C$ are a closed set and are as a class of densities totally bounded. 
Why is this? I cannot find a proof of it.
 A: Some context: a density means a nonnegative integrable function on $[0,1]$ with integral equal to $1$. Total boundedness is understood in the $L^1$ norm. 
Let $\mathcal F_C$ be the set of $C$-Lipschitz densities. I'll prove that it is totally bounded in the uniform norm $\sup|f|$, which will imply total boundedness in the $L^1$ norm (since the $L^1$ norm does not exceed the uniform norm.) 


*

*For every $f\in \mathcal F_C$ and every $x\in [0,1]$ we have $0\le f(x)\le C+1$. The lower bound is by definition of density. If the upper bound fails, that is $f(x)> C+1$ for some $x\in [0,1]$, then the Lipschitz condition implies $f(t)>1$ for all $t\in [0,1]$. But then $\int_0^1  f(t)\,dt>1$, a contradiction. 

*The family $\mathcal F_C$ is equicontinuous. Indeed, we can take $\delta=\epsilon/C$ in the definition of equicontinuity. 
From 1 and 2, using the Arzelà–Ascoli theorem, we conclude that $\mathcal F_C$ is compact with respect to the uniform norm, and therefore is totally bounded. 
A: Let $\mathcal{F}$ be the class of Lipschitz densities with Lipschitz constant bounded by $C>0$. Topologically, we regard $\mathcal{F}$ as a subspace of $L^{1}[0,1]$.
We first show that $\mathcal{F}$ is closed in $L^{1}$. Suppose $(f_{n})_{n=1}^{\infty}\subset\mathcal{F}$ converges to $f\in L^{1}$. Passing to a subsequence if necessary, we may assume by Riesz-Fischer that $f_{n}(x)\rightarrow f(x)$ almost everywhere (a.e.). So for a.e. $x,y\in[0,1]$,
$$\left|f(x)-f(y)\right|=\lim_{n\rightarrow\infty}\left|f_{n}(x)-f_{n}(y)\right|\leq\lim_{n\rightarrow\infty}C\left|x-y\right|=C\left|x-y\right|$$
Modifying $f$ on a set of measure zero if necessary, we may assume that $f$ is Lipschitz on $[0,1]$.
We now show that $\mathcal{F}$ is totally bounded in $L^{1}$. I claim that $\mathcal{F}$ is uniformly bounded by $1+C$. Indeed, by Weierstrass's Extreme Value Theorem, any $f\in\mathcal{F}$ attains its minimum and maximum at points $x_{\min},x_{\max}\in[0,1]$, respectively. Therefore
$$\sup_{0\leq x,y\leq 1}\left|f(x)-f(y)\right|\leq\left|f(x_{\max})-f(x_{\min})\right|\leq C$$
Since $\int_{0}^{1}f=1$, $0\leq f(x_{\min})\leq 1$, whence $f(x_{\max})\leq 1+C$. Define $M:=\sup_{f\in\mathcal{F}}f(x_{\max})$ (where $x_{\max}$ depends on $f$).
For integer $n\geq 1$, consider the step functions of the form
$$f_{i_{1}\cdots i_{n}}:=\dfrac{i_{1}M}{n}\chi_{[0,M/n]}+\dfrac{i_{2}M}{n}\chi_{[M/n,2M/n]}+\cdots+\dfrac{i_{n}M}{n}\chi_{[(n-1)M/n,M]}$$
where $i_{k}\in\left\{0,1,\ldots,n\right\}$ for $1\leq k\leq n$. Let $f\in\mathcal{F}$. On the interval $[jM/n,(j+1)M/n]$, let $i_{j}$ be such that $i_{j}M/n\leq f(jM/n)<(i_{j}+1)M/n$. Then
$\begin{align*}
\left|f(x)-\dfrac{i_{j}M}{n}\right|&\leq\left|f(x)-f\left(\dfrac{jM}{n}\right)\right|+\left|f\left(\dfrac{jM}{n}\right)-\dfrac{i_{j}M}{n}\right|\\
&\leq C\dfrac{M}{n}+\dfrac{M}{n}\\
&=(C+1)\dfrac{M}{n}
\end{align*}$
Hence,
$$\displaystyle\int_{[\frac{jM}{n},\frac{(j+1)M}{n}]}\left|f-\frac{i_{j}M}{n}\right|\leq(C+1)\dfrac{M^{2}}{n^{2}}$$
For $\varepsilon>0$ given, we can take $n$ sufficiently large so that the $(C+1)M^{2}/n<\varepsilon$, from which it follows that the open balls of radius $\varepsilon$ centered at the functions $f_{i_{1}\cdots i_{n}}$ totally bound $\mathcal{F}$.
