It is not possible to write a function representing the uniform distribution, say $\mathcal{U}$, on the real line, $\mathbb{R}$ (in a Bayesian context, this is an improper prior). My question is: is it possible to write a generalized function (aka, a distribution, but I avoid this term as it may cause confusion with the "distribution" used in probability terminology) corresponding to $\mathcal{U}(\mathbb{R}) $?
Additional text to give more context to the question. Just as Dirac's Delta can be represented as a limiting process of gaussians, i.e., $$ \delta(t) = \lim_{\sigma^2 \rightarrow 0} \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2 \sigma^2} t^2 } $$ in principle, $\mathcal{U}(\mathbb{R})$ could be represented as $$ \mathcal{K}(t) = \lim_{ \sigma^2 \rightarrow +\infty} \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2 \sigma^2} t^2 } $$ in some sense of the limit. Clearly, it must hold that $$ \int^{+\infty}_{-\infty} \mathcal{K}(t) dt = 1 $$ just as with Dirac's Delta. I do not know if there are technical difficulties that arise in measure theory for this case.
Related question: link. But I would like a bit more details as an answer.
Since no one answered, I would like to post here some useful insights on the matter.
The first, and most important, insight is that the improper uniform is inherently non-local. Distributions like Dirac's Delta, $\delta(t)$, have a local behavior, which is important. This issue makes the mathematical formalization of $\mathcal{K}(t)$ harder than it may appear.
The second comment is an attempted definition is the following. Consider the so-called integral mean (or average) over an interval $[a,b]$, given by $$ \frac{1}{|a-b|} \int^b_{a} f(t) dt $$ which is well-defined as long as the function $f$ is well-behaved. We can consider the limit of a sequence of increasing symmetric intervals as the mean of the function itself, given by $$ \mathbb{E}[f] \triangleq \lim_{N \rightarrow +\infty} \frac{1}{2N} \int^N_{-N} f(t)dt $$ and use this as our definition. In other words, the improper uniform can be thought of as an operator associating a real number to a function, i.e., $\mathbb{E} : \mathcal{F} \rightarrow \mathbb{R}$, where $\mathcal{F}$ is some function space with some regular properties (e.g. Schwartz space). This approach can likely be refined significantly, but I invite the author to post his answer since, as of now, no better definition has been given (unfortunately, the bounty expired).
There is, however, one significant aspect which I think can be improved upon. Intuition tells us that $\mathcal{K}(t)$ should be an infinitesimally thin "carpet", spread over $\mathbb{R}$. So, it makes sense that it should share some properties of the zero-function (the pointwise limit of gaussians for $\sigma^2 -> +\infty$). This can be accomplished via the following definition $$ \int_{\mathbb{R}} \mathcal{K}(t) f(t) dt = \lim_{\sigma^2 \rightarrow +\infty} \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{\mathbb{R}} e^{-\frac{t^2}{2 \sigma^2}} f(t) dt $$ Let us have a deeper look at this. For $f(t)=1$, we have that the result is $1$, so that $$ \int_{\mathbb{R}} \mathcal{K}(t)dt = 1 $$ is satisfied. Consider now $f(t)=t$, for which the integral on the LHS is zero. This is, intuitively, correct, as $\mathcal{K}(t)$ should behave like the zero-function and simply make the integral vanish, but not always! We should expect some functions, like $f(t)=c$, for any $c \neq 0$, not to be annihilated. Functions with divergences also may be into this category. In other words, the integral is zero for some (but not all!) functions $f$, making $\mathcal{K}(t)$ "look like" the zero-function. To be more precise, there exists a function space $\mathcal{F}_0 \subsetneq \mathcal{F}$, where $\mathcal{F} = \mathcal{F}(\mathbb{R})$ is the space of all functions in $\mathbb{R}$ (regular enough to have an integral), such that $$ \forall f \in \mathcal{F}_0: \quad \int_{\mathbb{R}} \mathcal{K}(t) f(t)dt = 0 $$ which means that $\mathcal{K}(t)$ acts like the zero-measure in $\mathcal{F}_0$. But, and this is the key point, it acts nontrivially when other functions ($\not\in \mathcal{F}_0$) come into play.