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In chapter 1.5 of Shreve & Karatzas's book, after proving that for continuous, square-integrable martingales, variations greater than 0 vanish and lower variations explode, it proceeds to argue that for this reason, they are not differentiable and we cannot define integrals of the form $\int_0^t Y_{s}(\omega)dX_{s}(\omega)$ pathwise. Question: why can't we define Lebesgue-Stieltjes integrals for integrators of unbounded first variation?

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    Forget for a comment that you have a stochastic integral and suppose that $F$ is a function in say $[0,\infty)$ that has not local finite variation, it for some interval $[a, b]$, $V(F;[a,b])=0$. How do you define a Radon measure from $F$? In Lebesgue integration, the minimal condition for the existence of a Radon measure associated to $F$ is that $F$ is of local bounded variation. Consider for example $f(x)=\frac{1}{|x-1|}$. Thus will produce a finite additive function $\mu_f$ in $\mathscr{B}([0,\infty)$ but not a Radon measure (the measure of the compact set ${0}$ is $\infty$. – Mittens Apr 10 '22 at 19:38
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    Finally, notice that if $X$ is continuous and of quadratic local finite variation, then then $X$ is of infinite local finite variation. (see here) – Mittens Apr 10 '22 at 19:44

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I think I have found the answer. I will follow Exercise 2.21 in chapter 3.2 of Yor & Revuz.

Let $X$ be a real-valued function on $[0,1]$. Let $\Delta = \{t_{1}\ldots t_{n}, 0\leq t_{1} \leq \ldots \leq t_{n}\leq n\}$ be a partition. For each $h \in C([0,1],\mathbb{R})$, define

$S_{\Delta}(h) = \sum_{t_{i}\in \Delta}h(t_{i})(X_{t_{i+1}}-X_{t{i}})$.

It is obvious that the map $h\rightarrow S_{\Delta}(h)$ is a bounded linear functional on $C([0,1],\mathbb{R})$, with norm $||S_{\Delta}||:=\sum_{t_{i}\in \Delta}|X_{t_{i+1}}-X_{t{i}}|$.

The Riemann-style integral is defined only if $\{S_{\Delta_{n}}(h)\}$ converges as $|\Delta_{n}|\rightarrow0$, for every $h \in C([0,1],\mathbb{R})$. By the Principle of Uniform Boundedness (also called the Banach-Steinhaus theorem), this implies that $\sup_{n} ||S_{\Delta_{n}}|| < \infty$. But $\sup_{n} ||S_{\Delta_{n}}||$ is exactly the first variation of $X$, provided that it exists. Hence, $X$ must be of bounded variation in order that the integral exists.

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