Is Dirac's delta function well-defined at Lebesgue points? 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 $L^1.$
 A: It seems to me that one can easily extend the definition so that, for each $f\in L^1$, for almost all $a$, $\int\delta(x-a)f(x)\,dx$ is defined; just define it to be the limit in the Lebesgue density theorem. But you seem to want the quantifiers in the other order: For almost all $a$, for all $f\in L^1$, $\dots$. I see no reasonable way to get this.  That is, "almost all" refers to an exception set of measure zero, and I see no way to make that set independent of $f$.
A: (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.
A: An element of $L^{1}$ is an equivalence class of functions which are equal a.e.. Elements of $L^{1}$ don't have pointwise values. Point values can make sense for specific cases, such as when there is a function in the equivalence class which happens to be continuous--that's because if there is such a function in the equivalence class, then there cannot be a different continuous function in the same equivalence class. However, point values are not determined in general.
