Reading through Griffiths 'Intro to Electrodynamics', he puts forth an argument for why the potential of the electric field $V$ is continuous across the boundary of any surface charge. I'll put it forth here briefly :
The definition of the electric potential at a point $r$ is $V(r) = -\int_{O}^{r} \overrightarrow{E} \cdot dl$ , (due note that with some derivation, this leads to the equivalent definition $\overrightarrow{E} = \nabla V)$ where $O$ is an arbitrary reference point (usually chosen as $\infty $). He then says to evaluate if there is a difference/discontinuity of this function across a surface that holds a charge, we should evaluate the difference right above and right below a surface at the points $a$ and $b$
By definition $V(b) - V(a) = -\int_{O}^{b} \overrightarrow{E} \cdot dl - (-\int_{O}^{a} \overrightarrow{E} \cdot dl)$
$ = \int_{b}^{O} \overrightarrow{E} \cdot dl + \int_{O}^{a} \overrightarrow{E} \cdot dl = \int_{b}^{a} \overrightarrow{E} \cdot dl$
$= -\int_{a}^{b} \overrightarrow{E} \cdot dl $
this is where my question comes in, he proceeds to argue that in order to evaluate the difference in the potential function right above and right below this above pictured surface, we make $a$ and $b$ closer and closer. He doesn't frame it this way but this is how I interpreted this mathematically : Re-express $V(a)$ and $V(b)$ as infinitesimal points right above and below the point a point lying on the surface called $s$. As in
$$V(below) = V(a) = \lim_{\epsilon \rightarrow 0} V(s - \epsilon)$$ and $$ V(above) = V(b) = \lim_{\epsilon \rightarrow 0} V(s + \epsilon)$$
then to evaluate the difference right above and below the surface, we do an equivalent derivation as we did for $V(b) - V(a)$ :
$$ V(above) - V(below) = \lim_{\epsilon \rightarrow 0} [V(s + \epsilon) - V(s-\epsilon)]$$ $$ = \lim_{\epsilon \rightarrow 0} [-\int_{s - \epsilon }^{s+ \epsilon} \overrightarrow{E} \cdot dl] = 0 $$
he explains my above understanding in english : "as the path length shrinks to zero, so too does the integral" leaving us with $V(above) - V(below) = 0 \rightarrow V(above) = V(below)$ , and thus the potential function as always continuous across any boundary.
This seems like mathematical trickery however, surely you could apply this argument to any function that can be expressed in terms of a line integral, continuous or not, and say "well it's continuous everywhere because if I evaluate the difference of this function at any 2 points where I want to check for a discontinuity, just take the line integral, shrink the length to 0 and then so too does the difference, thus the function is continuous".
This next part is non-essential to the question, but I tried framing the argument differently in 1D, however I'm unsure of how sound what I've done is and I'd love some input :
By expressing the electric field along a line in terms of a scalar function, as in imagine we had an electric field like so :
Only consider the electric field lying on the green line. Since the situation is simple on this line, we do not require to represent the electric field with anything more than a scalar quantity at every $ $$z$ value. If we think of positive values as representing an electric field vector pointing upward on the $z$-axis, and negative values as pointing downward on the $z$-axis, then the electric field on this line can be represented simply as $E(z) : \mathbb{R} \rightarrow \mathbb{R}$. Then we create an equivalent 'definition' for 1D where instead of a line integral, the potential is a single variable integral $V(z) = -\int_{O}^{z} \overrightarrow{E}dz'$. Then once again similar to Griffiths the difference between any 2 points on this '1D conversion' potential would be $V(b) - V(a) = - \int_{O}^{b} \overrightarrow{E}dz' - (-\int_{O}^{a} \overrightarrow{E}dz') = - \int_{a}^{b} \overrightarrow{E}dz'$
Applying the next part of Griffiths argument doesn't make sense here either, if I want to evaluate the potential right above and right below a surface as Griffiths does for our new 1D analogy, I'd do similar to my above epsilon argument and say
$$ V_{1D}(above) - V_{1D}(below) = \lim_{\epsilon \rightarrow 0} [V_{1D}(s + \epsilon) - V_{1D}(s-\epsilon)]$$ $$ = \lim_{\epsilon \rightarrow 0} [-\int_{s - \epsilon }^{s+ \epsilon} \overrightarrow{E}dz'] = 0 $$
Again this comes off as mathematical trickery to me, and I don't see why it wouldn't apply for a completely discontinuous jump function at $x=3$ like such
I’m not even sure if the above function at $x=3$ is integrable, but how would I catch that discontinuity/lack of integrability mathematically from Griffiths’ argument?
Any help/insight would be appreciated
By $\int_0^a E.dl$ i believe you mean the path $l$ is along a single axis (say) z-axis. Assuming this, lets say $E=\delta(z-s)$ then $\lim_{\epsilon \rightarrow 0}\int_{s-\epsilon}^{s+\epsilon} E.dl = \lim_{\epsilon \rightarrow 0}\int_{s-\epsilon}^{s+\epsilon} E dz = 1$. This is what he is refering to as discontinuity. In general if there is a jump or discontinuity in $E$ at $s$, the integral $\lim_{\epsilon \rightarrow 0}\int_{s-\epsilon}^{s+\epsilon} E dz$ need not be $0$.
Intuitively if $V = f + \sum_i a_i u(z-s_i)$ where $f$ is continuous and $u$ is a unit step function, then $V$ has disontinuities at $\{s_i\}$ and at these points the derivative is $\delta(.)$ which was there in the expression for $E$ and your integral does not vanish because these $\delta(.)$ functions are of unit area and its support occupy vanishingly miniscule length in $z$-axis.
So the integral vanishes if $E$ is continuous or bounded at $s$. If he says the integral is $0$, it means $E$ is continuous or bounded at $s$. Note that i did not say "iff" E is continuous or bounded.
The moral of the story is at discontinuities the derivatives of the functions are written or expressed as $\delta(.)$ function and the value of the function at these points is found by pretending that you integrate around the discontinuity instead of evaluating the $\delta(.)$ function simply because of mathematical convenience as discontinuity implies derivative does not exist in normal sense. So using $\delta(.)$ we express derivative in a certain different sense called distributional sense.