Which functions can be approximated by Hölder continuous functions in a "bounded way"? I'm interested in the space of functions $f: \mathbb{R}^d \to \mathbb{R}$ satisfying the following property:

There exist a sequence $(\alpha_n)_{n \in \mathbb{N}} \subseteq (0,1]$ and functions $f_n$ which are Hölder continuous with Hölder exponent $\alpha_n$ such that $$f_n(x) \xrightarrow[]{n \to \infty} f(x) \quad \text{for (Lebesgue)almost all $x \in \mathbb{R}^d$} \tag{1}$$ and $$\sup_{n \geq 1} \|f_n\|_{\alpha_n} < \infty. \tag{2}$$

Here, $\|\cdot\|_{\alpha}$ denotes the $\alpha$-Hölder norm, i.e. $$\|g\|_{\alpha} := \sup_x |g(x)| + \sup_{x \neq y} \frac{|g(x)-g(y)|}{|x-y|^{\alpha}}.$$
Edit: Note that $\alpha_n$ might tend to $0$, and therefore $(2)$ does not imply $$\sup_{n \geq 1} \|f_n\|_{\alpha} < \infty$$ for some $\alpha>0$.

Obviously, any function $f$ which is Hölder continuous has the above property. The interesting things happen if $\alpha_n \to 0$ for $n \to \infty$. For instance, there are discontinuous functions which can be approximated in the above sense; e.g. the Heaviside function
$$f(x) = 1_{(0,\infty)}(x)$$
can be approximated by
$$f_n(x) := 1_{(0,\infty)}(x) \cdot \min\{|x|^{1/n},1\}.$$
I'm aware of the fact that Sobolev functions can be approximated by Hölder continuous mappings, but the results, which I know, do not provide a uniform bound $(2)$.
I'm wondering how "big" the above defined space of functions is, i.e. which known function spaces are (not) contained in it. I would be very happy about references and your thoughts on the problem.
 A: Edit: At first I read the question wrong, assuming that $||f_n||_\alpha$ was supposed to be bounded, not just $||f_n||_{\alpha_n}$. First a comment on the actual problem (not certain if it's a solution to what the OP actually has in mind because I'm not clear on a certain quantifier, then the original answer, for the case $\alpha_n=\alpha$.
The condition "$||f_n||_{\alpha_n}$ bounded" seems a little curious. For example:


Curious Lemma: Suppose that $||f||_\infty\le 1$ and there exists $c$ such that $|f(x)-f(y)|\le c|x-y|$. Then there exists $\alpha>0$ with $||f||_\alpha\le 4$.


Proof: Choose $\alpha\in(0,1)$ so that $2c^\alpha<3$. If $|x-y|>1/c$ then $$\frac{|f(x)-f(y)|}{|x-y|^\alpha}\le\frac2{c^{-\alpha}}\le3.$$On the other hand if $|x-y|<1/c$ then
$$\frac{|f(x)-f(y)|}{|x-y|^\alpha}\le\frac{c|x-y|}{|x-y|^\alpha}
=c|x-y|^{1-\alpha}\le c^{\alpha}\le 3.$$
That really seems like it can't be right. But I don't see the error. If we're being careful  we note that in the first case $\alpha>0$ implies $|x-y|^\alpha>(1/c)^\alpha$, while in the second case $1-\alpha>0$ implies $|x-y|^{1-\alpha}<(1/c)^{1-\alpha}$.


Curious Corollary. If $f$ is measurable and $||f||_\infty\le1$ then there exist $f_n$ and $\alpha_n>0$ such that $f_n\to f$ almost everywhere and $||f_n||_{\alpha_n}\le 4$.


(Of course the converse is trivial, since $||f_n||_\infty\le||f_n||_{\alpha_n}$.)
Proof. Let $\phi_n\in C^\infty_c(\mathbb R)$ be an approximate identity, with $\phi_n\ge0$ and $\int\phi_n=1$ as usual. Let $f_n=\phi_n*f$. Then $f_n\to f$ almost everywhere. Also $||f_n||_\infty\le||f||_\infty||\phi_n||_{L^1}\le 1$ and $||f_n'||_\infty
\le||f_n||_\infty||\phi_n'||_{L^1}<\infty$. Curiously, the size of $||f_n'||_\infty$ doesn't matter, the lemma gives our $\alpha_n$. 

Original, assuming $\alpha_n=\alpha$:


Easy Theorem $f$ can be approximated in this sense if and only if there exists a Holder-continuous $g$ with $f=g$ almost everywhere.


For the not entirely trivial direction: Suppose that $f_n\to f$ almost everywhere and $||f_n||_\alpha$ is bounded. Arzela-Ascoli shows that some subsequence converges uniformly on compact sets to $g$. So $g$ is Holder continuous and $f=g$ almost everywhere.
(Or without Arzela-Ascoli: Say $E$ is a set of full measure and $f_n\to f$ pointwise on $E$. Then $f|_E$ satisfies a Holder condition. Hence $f|_E$ is uniformly continuous; since $E$ is dense $f|_E$ has a unique extensiion to a continuous function $g:\mathbb R^n\to\mathbb R$, which is also Holder continuous.)

Thoughts When people talk about $f$ being a limit of functions with some sort of uniform smoothness property the point is typically to see what sort of smoothness this implies about $f$; the Easy Theorem is a typical example. So typical that when I first read the question I assumed it was asking about the Easy Theorem.
Then when I saw the question was actually about $||f_n||_{\alpha_n}=O(1)$ it seemed a little fishy; that really didn't seem like a "uniform" smoothness condition. And sure enough, the Curious Corollary shows that $||f_n||_{\alpha_N}=O(1)$ gives no more smoothness than just $||f_n||_\infty=O(1)$.
Anyway, the point to this section is to suggest that if one is interested in what smoothness on $f$ follows from bounds on $||f_n||_{\alpha_n}$ one might obtain more interesting results from assuming that $||f_n||_{\alpha_n}\to0$ at some rate, depending on $(\alpha_n)$.
Except of course that's not what I mean; it's clear from the definition that if $||f_n||_{\alpha_n}\to0$ then $f=0$. We defined $||f||_\alpha$ the way we did just because it's nice to have a norm. Instead define a seminorm $$\rho_\alpha(f)=\sup_{x\ne y}\frac{|f(x)-f(y)|}{|x-y|^\alpha}.$$
Then it seems to me that if one is interested in all this one might want to consider how smooth $f$ must be if $||f_n||_\infty=O(1)$ and $$\rho_{\alpha_n}(f_n)\le ch(\alpha_n)$$for some function $h$ with $h(\alpha)\to0$. For example $h(\alpha)=\alpha^\beta$ springs to mind...
A: But doesn't $\alpha_n\to 0$?
I would say also that any bounded $f$ can be approximated in this way.
1) $f$ bounded (say, $|f|\le C$) can be approximated a.e. by continuous functions $f_n(x)$ (with $|f_n|\le C$).
2) $f_{n,L}(x):=\min_y f_n(y)+L|x-y|$ is a $L$-Lipschitz approximation of $f_n$ (with $f_{n,L}\to f_n$ uniformly as  $L\to\infty$ and $|f_{n,L}|\le C$).
3) one has
$$ \frac{|f_{n,L}(x)-f_{n,L}(y)|}{|x-y|^\alpha} \le \min\Big\{ L|x-y|^{1-\alpha}, \frac{2C}{|x-y|^\alpha}\Big\}. $$
This is maximal for $|x-y|=2C/L$, with value $2C(L/(2C))^\alpha$. Choosing $\alpha= \log 2/(\log(L/(2C))$ (which goes to $0$ as $L\to\infty$) gives the value $4C$.
