Can a non Lebesgue integrable function tend to a $\delta$ distribution? Consider this definition of $\delta (x)$
$$\int_\mathbb{R} \delta(x)f(x)\mathrm{d}x=f(0) $$
I saw a question asking to prove that $$ \frac{1}{\pi x}\sin\left(\frac{\pi x}{\varepsilon}\right)$$
approaches $\delta (x)$ as $\varepsilon$ goes to $0$. To try to answer I plugged it into the definition
$$ \lim\limits_{\varepsilon\rightarrow0}\int_\mathbb{R}\frac{1}{\pi x}\sin\left(\frac{\pi x}{\varepsilon}\right)f(x)\mathrm{d}x=\lim\limits_{\varepsilon\rightarrow0}\int_\mathbb{R}\frac{1}{\pi t}\sin(t)f\left(\frac{\varepsilon t }{\pi}\right)\mathrm{d}x=$$
Now, even assuming we could permutate limit and integral (which I don't think we can) we would get
$$ f(0)\frac{1}{\pi}\int_\mathbb{R}\frac{\sin(t)}{t}\mathrm{d}t$$
Now, this is improperly Riemann integrable, and the result would correctly be $f(0)$. Nevertheless, it is not Lebesgue integrable.
My question is: can we really say that this tends to a $\delta$? And secondly, can we permutate integral and limit?
 A: To understand the convergence of (test and generalized) functions to the delta distribution, one must talk about the space of such functions about its norm.
The discussion starts with the Schwartz space over the real line: $\mathcal{S}(\mathbb{R})$, and its dual space. If you are unfamiliar with this space, think of very well-behaved functions ($C^\infty$ and all the derivatives are integrable). One then realizes that the dual space, that is the space of functionals from $\mathcal{S}(\mathbb{R})$ to $\mathbb{R}$, contains $L^\infty$.
Answering your specific question, any $L^\infty$ function is a test function (like the Dirac distribution), and so terms like "distance" and "convergence" are well defined, without going into details as to how the norm is defined. Furthermore, the function $\sin(\pi t/\varepsilon)/\pi t \in L^\infty$ does indeed converge to the Dirac distribution, but I do not know an easy way to show this, so I'll sketch what I'm thinking:
Suppose $f$ is a function in the Schwartz space, then first notice the integral of $f(t) \cdot \sin(t/\varepsilon)/t$ exists.


*

*$f$ must be bounded at the tail by ever decreasing numbers $1, 1/2, 1/4 \ldots$. Say $f$ is bounded by $1$ in the positive part of the real line in the intervals $[a_1, 2_1]$, and by $1/2$ in $[a_2 , a_3]$ etc. with $0<a_{n-1} < a_n$ and $a_n \rightarrow \infty$. We do the same in the negative part of the real line with with the interval $[b_2,b_1]$ for $1$, and $[b_3,b_2]$ for $1/2$, with the endpoints determined by the rule $b_n<b_{n-1}<b_1 <0$ and $b_n \rightarrow -\infty$.

*In each of these line segments the integral vanishes (this is the Riemann-Lebesgue lemma), and we are left to deal with the integral in the finite interval $[b_1,a_1]$.

*The problem is now reduced to an integral over a finite interval and I leave this to you. If you are unfamiliar with how to show this from here; search for Approximation of unity.

A: Let $f$ be a test function, and suppose $f$ is supported on $[-a,a]$ for some $a > 0$. Since $f$ is uniformly continuous, given $\eta > 0$, there corresponds a $\delta > 0$ such that for all $x,y$, $\lvert x - y\rvert < \delta$ implies $\lvert f(x) - f(y)\rvert < \eta$. Let $0 < \epsilon < \dfrac{\pi \delta}{\lvert a\rvert}$. For all $t\in [-a,a]$, $\lvert f(\frac{\epsilon t}{\pi}) - f(0)\rvert < \eta$. It follows that  
$$\frac{1}{\pi}\int_{-\infty}^\infty \frac{\sin t}{t}\left[f\Bigl(\frac{\epsilon t}{\pi}\Bigr) - f(0)\right]\, dt$$
is bounded by $C\eta$, where 
$$C = \frac{1}{\pi} \int_{-a}^a \left\lvert \frac{\sin t}{t}\right\rvert\, dt$$
Since $\eta$ was arbitrary, 
$$\lim_{\epsilon\to 0} \int_{-\infty}^\infty \frac{1}{\pi x}\sin\left(\frac{\pi x}{\epsilon}\right) f(x)\, dx = \lim_{\epsilon \to 0} \frac{1}{\pi}\int_{-\infty}^\infty \frac{\sin t}{t} f\left(\frac{\epsilon t}{\pi}\right)\, dt = \frac{1}{\pi} \int_{-\infty}^\infty \frac{\sin t}{t}f(0)\, dt = f(0)$$
and thus $\dfrac{1}{\pi x}\sin\left(\dfrac{\pi x}{\epsilon}\right)$ converges to $\delta(x)$ in the distributional sense as $\epsilon \to 0$.
