TITLE: Is Dirac's delta function well-defined at Lebesgue points?
QUESTION [6 upvotes]: Usually in textbooks, 
$$\int_{\mathbb{R}^d} \delta(\mathbf{x}-\mathbf{y})f(\mathbf{x}) = f(\mathbf{y})$$
holds given $f$ is continuous. On the other hand, the definition of Lebesugue point $\mathbf{y}$ is that the following limit exists
$$
\lim_{\epsilon\rightarrow0}\frac{1}{|B_{\epsilon}|}\int_{B_{\epsilon}}f(\mathbf{x}),
$$
where $B_{\epsilon}$ is the ball of radius $\epsilon$ centered around $\mathbf{y}.$
Lebesgure differentiation theorem says for $L^1(\mathbb{R}^d)$ function almost every point is a Lebesgue point. 
All these facts let me think the delta function is also well-defined almost everywhere for $f\in L^1(\mathbb{R}^d)$ because the definition of Lebesgure point is exactly an example of approximation of the delta function. More precisely, can the definition of the delta functional be extended to the domain $L^1$ in this way?
For delta function, we take this definition: The delta function is a generalized function that can be defined as the limit of a class of delta sequences. My question is to extend this functional to $L^1$ such that the extended delta function is a linear bounded functional on $L^1.$

REPLY [5 votes]: (Too long for a comment)
In my opinion, this is an interesting curiosity and I think the correct answer is what Andreas has said - simply define $\langle \delta_a,f\rangle$ as the Lebesgue limit.  This might save you some space if you need to say something like "the value of $f$ at $x$" in the context of an $L^1$ function (for which function values are not usually uniquely defined, as pointed out by T.A.E.). However I can't see this being of much use beyond that context, at least as far as distribution theory goes. The whole point of the theory of distributions is to work with as nice of a test function space as possible so that you can deal with 'nastier' distributions.  In this sense $L^1$ (or any space defined in terms of integrability properties) makes a very poor test function space because it's too 'big'.  Indeed most of the time it works precisely the other way around - $L^1_{loc}$ serves as our space of 'regular' distributions, i.e. we test with $L^1$ functions, not on $L^1$ functions.