I struggle to link the concepts I learnt in measure theory and the one I learnt in analysis talking about antiderivative.
Especially, this equality disturbs me (not in the sense that I don't agree with it, but just that it seems to be contradicting antiderivatives) :
$$ f(x_0) = \int f(x) d( \delta_{x_0} ) $$
Are antiderivatives particularly specific to analysis over classical fields like $\mathbb R^n$ and $\mathbb C^n $?
Is it still true that in general an integral is continuous ? That was a given in my head and I have the impression that in fact it is not true in general...
To make that last sentence clearer :
I know that $$f(x) = \int^x_{-\infty} g(y) dy $$ are continuous functions.
Depending on the continuity of $g$ wrt $z$, $$f(z) = \int g(x,z) dx $$ is also continuous. What about the case where the variable appears in the measure?
I believe the question of antiderivative and continuity are linked. Please let me know if it is unclear.
I'm not entirely sure what you mean when you ask if "an integral is continuous". Do you mean that, given a regular measure $\mu$ on $\mathbb{R}$ and $x_0\in\mathbb{R}$, the function $x\mapsto \mu([x_0,x])$ is continuous on $[x_0,\infty)$? This is not true (e.g. take the point-mass $\delta_1$ and $x_0=0$).
Now on to you question.
In principle, measures are simply a notion of "size" on a set. Nothing about derivatives.
So let us consider a probability space $(X,\mu)$. Could be $[0,1]$ with the usual Lebesgue measure, for example. More generally, one common interpretation is that $X$ is a set of possible states of some physical system (e.g. could be $\mathbb{S}^2\times[0,1]$, whose elements $(x,t)$ have a point $x$ in a sphere/the earth and a point $t$ in time).
Then consider some positive $L^1$ function $f$ on $X$. This is just something that we can measure on our system (e.g. $f(x,t)$ is the temperature of point $x$ of the sphere at time $t$). We can consider the function $$F\colon A\mapsto\int_A f(x)d\mu(x)$$ defined for measurable sets $A$. This gives us "how much of whatever $f$ measures is there in $A$".
It just so happens that the map above is also a measure on $X$. It can also be seen as a "weighted version of $\mu$ by $f$". Since the set "$A$" is the argument of the function $F$, we can give it a better name by just ommiting $A$ in the formula: Thus $$\int fd\mu$$ is a measure, which takes a set $A$ to $\int_A f(x)d\mu(x)$.
We can go in the opposite direction via the Radon-Nikodym theorem: If $\nu$ and $\mu$ are nice enough measures on $\nu$ and $\nu\ll\mu$, then there exists a function $f$ on $X$ such that $\nu=\int fd\mu$. This function $f$ is unique (up to null sets), and is denoted by $\frac{d\nu}{d\mu}$. It is called the Radon-Nikodym derivative of $\nu$ wrt $\mu$.
The following formulas are all valid, as long as everything is sufficiently regular:
so Radon-Nikodym derivatives are really the opposite of integrals.
Ok. So now we have a new version of derivatives, which have nothing to do, it seems, with the derivatives in Calculus. Let's see what we can do:
Consider a sufficiently regular measure $\mu$ on $\mathbb{R}$. Let us associate to $\mu$ the function $$F_\mu(x)=\begin{cases}\mu(0,x]&\text{ if }x>0\\-\mu(x,0]&\text{ if }x<0\end{cases}$$ The measure $\mu$ is completely determined by the function $F_\mu$ (in the sense that $F_\mu=F_{\mu'}$ implies $\mu=\mu'$).
Nice. So (sufficiently regular) measures on $\mathbb{R}$ are functions.
Let's do that with the simplest measures we can think of: Let $f\colon\mathbb{R}\to\mathbb{R}$ be continuous, and take $\lambda$ the Lebesgue measure on $\mathbb{R}$. Construct the measure $\int fd\lambda$. By the discussion above, this measure is determined by the function $$F(x)=\int_0^x f(t)dt$$
But we have seen above that the Radon-Nikodym derivative of $\int fd\lambda$ is $f$. How does that translate to the function $F$? The answer is given by the Fundamental Theorem of Calculus: