No, you cannot disavow Lebesgue.

Lebesgue integration (and measure theory) are indispensable in stochastic calculus, especially for finance applications.

Take for example the martingale representation theorem, which is central to the theory of dynamic hedging and replication. This version is essentially Rogers and Williams (1987), Theorem 36.5:

**Theorem**

*The martingale representation theorem*

Let \(W\) be a standard Brownian motion of dimension \(K\). If \(X\) is a martingale with respect to the augmented filtration generated by \(W\), then there exists a process \(b \in \mathcal{L}^{2}\) such that

$$X(t) -X(0) = \int_{0}^{t} b \, dW$$ (in the sense that the two processes are indistinguishable).

But what does it mean to say that the process \(b\) is in \(\mathcal{L}^{2}\)? It means that \(b\) is measurable and adapted and pathwise square integrable on bounded intervals. In other words,

$$\int_{0}^{t} \|b\|^{2} \, ds < \infty $$ with probability one, for all \(t \in [0,\infty)\). Or, to write it out in detail, if the underlying probability space is \((\Omega,\mathcal{F},P)\), then the requirement is that

$$\int_{0}^{t} \|b(\omega,s)\|^{2} \, ds < \infty $$ for \(P\)-almost all \(\omega \in \Omega\), for all \(t \in [0,\infty)\).

For each \(\omega\), the time integral is a Lebesgue integral. The martingale representation theorem does not in any way guarantee that the function \(\|b\|^{2}\) is pathwise Riemann integrable. How could it?

So the statement of the martingale representation theorem requires Lebesgue integration and measure theory.