If path integrals aren't well-defined, how can they have any physical meaning? I am confused about a particular point about the nature of path integration. According to what I've read, what we really mean when we say functional integration is
\begin{equation} 
\int\mathcal{D}\phi=
\int_{-\infty}^\infty\prod_x d\phi(x)
\end{equation} 
in the sense that
\begin{equation} 
\int\prod_{i=1}^n dx_i=\underbrace{\int\cdots\int}_n dx_1\cdots dx_n
\end{equation}
As I understand it, we integrate at each point over all field configurations at that point. 
But supposedly one never actually does a path integral because this is not well-defined. $\prod_x$ makes no sense. 
My Question:
Why doesn't $\prod_x$ make any sense?  How can the integral have any meaning if its not well-defined? 
 A: When I said we don't actually do path integrals, what I meant to say is that we can do some very specific path integrals and the way we do them is rather ad-hoc. In other owrds, it's highly nontrivial and not straightforward. To show this, I'll do a very general path integral for you. (I had most of this typed up already for another reason.)
The vacuum transition amplitude for a set of quantum fields collectively denoted by $\varphi(x)$
is given by the path integral
$$Z[0]:=\langle\text{VAC, out}|\text{VAC, in}\rangle=\int D\varphi\;\exp\left(i\int d^4x \,\mathcal{L}(\varphi)\right)$$
where $D\varphi$ is the path measure
$$D\varphi:=\prod_{x,\ell}d\varphi_\ell(x)$$
and $\ell$ runs over components and species. Let $A$ be a real symmetric $N\times N$ matrix and $x$ and $J$ be vectors with $N$ components. Then we have the following integral formula:
$$\iint\dotsb\int dx_1dx_2\dotsb dx_N\,\exp\big((i/2)x^\text{T}Ax+iJ\cdot x\big)=\sqrt{\frac{(2\pi i)^N}{\text{det} \,A}}\exp\big(-(i/2)J^\text{T}A^{-1}J\big)$$ 
Now let us add to the integral of the Lagrangian in the exponent a current term that we will write as the integral of $J\varphi$. For example, in electromagnetism this would be $A\cdot J$ with $A_\mu$ the photon field and $J_\mu$ the conserved 4-current. This defines the partition function, which is a functional of the current:
$$Z[J]:=\int D\varphi\;\exp\left(i\int d^4x \,\big[\mathcal{L}(\varphi)+J\varphi\big]\right)$$
Now we separate the Lagrangian into a free quadratic part $\mathcal{L}_0$ and a (possibly) more complicated interaction part $\mathcal{L}_1$:
$$\mathcal{L}(\varphi)=\mathcal{L}_0(\varphi)+\mathcal{L}_1(\varphi)$$
For our purposes we set the interaction to zero: we only want single-particle states$^1$. We write the integral of the quadratic part very generally as
$$\int d^4x\,\mathcal{L}_0=-\frac{1}{2}\int d^4x\,d^4y \sum_{\ell,m}\mathcal{D}_{\ell x,my}\varphi_\ell(x)\varphi_{m}(y)$$
and the source term as
$$\int d^4x \,J\varphi=\int d^4x\, \sum_\ell J^\ell(x)\varphi_\ell(x) $$
Plugging this into the partition function
$$Z[J]=\int D\varphi\;\exp\left(-\frac{i}{2}\int d^4x\,d^4y \sum_{\ell,m}\mathcal{D}_{\ell x,my}\varphi_\ell(x)\varphi_{m}(y)+i\int d^4x\, \sum_\ell J^\ell(x)\varphi_\ell(x)\right)$$
we note that if we think of the integrals as sums, we get a infinite multiple integral in the above form$^2$. Denote the inverse of $\mathcal{D}_{\ell x,my}$ by $\Delta_{\ell m}(x,y)$. They are inverses in the sense that
$$\sum_{m}\int d^4y \,\mathcal{D}_{\ell x,my}\Delta_{mn}(y,z)=\delta^4(x-z)\delta_{\ell n}$$
The matrix $\Delta$ is called the propagator. In the absence of external fields, translation invariance will make $\mathcal{D}$ necessarily only a function of $x-y$, which can be written as a Fourier integral
$$\mathcal{D}_{mx,ny}:=(2\pi)^{-4}\int d^4p\, e^{ip\cdot(x-y)}\mathcal{D}_{mn}(p)$$
and the inverse relation is then
$$\Delta_{mn}(x,y)=(2\pi)^{-4}\int d^4p\, e^{ip\cdot(x-y)}\mathcal{D}_{mn}^{-1}(p)$$
where $\mathcal{D}^{-1}(p)$ is the ordinary matrix inverse of $\mathcal{D}(p)$. We can now evaluate the $Z[J]$ path integral. The factor in the finite dimensional case generalizes as
$$\sqrt{\frac{(2\pi i)^N}{\text{det} \,A}}\longrightarrow \lim_{N\rightarrow\infty}\sqrt{\frac{(2\pi i)^N}{\text{det} \,\mathcal{D}}}$$
The numerator part diverges and obviously has no meaning$^3$. The denominator might appear more interesting, but it does not depend on the fields or the currents. We call this overall factor $\mathcal{C}$. The term in the exponent generalizes as
$$-(i/2)J^\text{T}A^{-1}J\longrightarrow \frac{i}{2}\sum_{mn}\int d^4x\,d^4y\, J^m(x)\Delta_{mn}(x,y)J^n(y)$$
(note the sign change because $\mathcal{D}$ is defined with an extra minus sign). Putting this all together, we have
$$Z[J]=\mathcal{C}\exp\left(\frac{i}{2}\sum_{mn}\int d^4x\,d^4y\, J^m(x)\Delta_{mn}(x,y)J^n(y)\right)=\mathcal{C}\exp\big(iW[J]\big)$$
where $W[J]$ is the quadratic functional 
$$W[J]:=\frac{1}{2}\sum_{mn}\int d^4x\,d^4y\, J^m(x)\Delta_{mn}(x,y)J^n(y)$$



*

*The quantization of the interaction terms leads to the vertices of QFT.

*Here $\mathcal{D}_{mx,ny}$ is treated as a matrix very loosely. What we should really do is introduce a lattice on spacetime such that the labels $x$ and $y$ are discrete.

*I think this term has meaning in certain situations, but I cannot recall any at the moment.
A: According to Faddeev and Slavnov (Gauge fields: Introduction to Quantum Theory), "...all those properties of the Feynman integral that are used in practice in the perturbation theory are derived directly from the definition of the quasi-Gaussian integral and can be rigorously established independent on the issue of existence of Feynman integral measure. Therefore, in the framework of perturbation theory, the formalism of functional integration is a quite rigorous method, and results obtained using this method do not require additional proof." (I quote  by a Russian edition).
