I'm confused on how to apply the law of total probability when the summation is $\sum_{0}^{1}$ i.e. from 0 to 1, including all possible decimal numbers in between, rather than a summation of $\sum_{a}^{b}$ where a and b are integers. Is the following the correct approach (using an integral from 0 to 1)?
If I apply the law of total probability to P(Y=1), does it equal:
$ \int_{0}^{1} $P(Y=1|X=x)P(X=x)dx
where Y $\in$ {0,1} and x$\in$[0,1]
No.
If e.g. $X$ has continuous distribution then your integrand is $0$ for every $x$ hence also the integral takes value $0$.
Go for:
$$\begin{aligned}P\left(Y=1\right) & =\mathbb{E}\left[\mathbf{1}_{Y=1}\right]\\ & =\mathbb{E}\left[\mathbb{E}\left[\mathbf{1}_{Y=1}\mid X\right]\right]\\ & =\int\mathbb{E}\left[\mathbf{1}_{Y=1}\mid X=x\right]F_{X}\left(x\right)\\ & =\int P\left[Y=1\mid X=x\right]F_{X}\left(x\right) \end{aligned} $$