Let $X_t$ be a solution of a SDE. Does the set $\{X_t \in \{p\}\}$ has null measure? This question was previously posted on https://math.stackexchange.com/questions/3981156/let-x-t-be-a-solution-of-a-sde-does-the-set-x-t-in-p-has-null-meas.
I think this question is easy. However, I have not been able to solve it.
Let $a,\sigma:\mathbb{R}\times \mathbb R\to\mathbb{R}$, smooth functions such that $\sigma>0$. Consider the 1-dimensional SDE,
$$dX_t = a(X_t,t) dt + \sigma(X_t,t) dW_t$$
$$X_0 = x_0\in\mathbb{R}. $$
where $W_t$ is the standard Brownian motion.
Fixing $y\in\mathbb R$ and $t>0$, I was interested in showing that$$\mathbb{P}\left(\{\omega \in \Omega;\ X_t = y\}\right)=0.$$
Where $(\Omega,\mathcal F, \mathbb P)$, is the probability space being considered.
Does anyone know if the above equation is true? A reference would be enough for me.
 A: With my washed-up SDE knowledge from option pricing, I came up with the following. I am not sure if it is correct reasoning, but I am open to discuss it so I can also learn the right answer If I am wrong.
We have:
$$X_t - X_{t-1} = a(X_{t-1}, t-1)(t - t + 1) + \sigma (X_{t-1}, t-1)(W_t - W_{t-1})$$
$$X_t = X_{t-1} + a(X_{t-1}, t-1)(t - t + 1) + \sigma (X_{t-1}, t-1)(W_t - W_{t-1})$$
From the properties of Brownian motion, we know that:
$$(W_t - W_{t-1}) \sim N(0, 1)$$
Also note that because we are at time $t$, $X_{t-1}$ is realized, thus $a(X_{t-1}, t-1)$ and $\sigma (X_{t-1}, t-1)$ is deterministic and the only random component is Brownian motion increment. Thus, distribition of $X_t$ is given by:
$$X_t \sim N(X_{t-1}+a(X_{t-1}, t-1), \sigma (X_{t-1}, t-1))$$
Since it is a normal distribution and normal distribution is a continuous distribution. And for all continuous distributions, we have:
$$P(X=x) = 0 \ \forall x$$
Thus:
$$P(X_t=y) = 0 \ \forall y$$
Looking forward to your feedback
Edit:
This is the explanation of what @stats_model commented.
A: You say that this question is 'easy' but I would say that it is not so easy, having had a quick look at the relevant literature. Possibly the most elementary and accessible answer to this question is contained in the paper by N. Fournier, J. Printems,
Absolute continuity of some one-dimensional processes,
Bernoulli, 16 (2010), pp. 343-360. Section 3 of the paper considers SDEs of the form $$X_t=x+\int_0^t \sigma(X_s)\kappa (s,(X_u)_{u\leq s} H_s) \text{d}B_s+\int_0^t b(s,(X_u)_{u\leq s},H_s)\text{d}s$$ for suitable functions $\sigma$, $\kappa$ and $b$ and an auxiliary process $(H_t)_{t\geq 0}$. Theorem 3.1 states that, given suitability nice coefficient functions, the law of $X_t$ has a density on $\{x\in\mathbb{R}: \sigma(x)\neq 0\}$ whenever $t>0$. I think that this theorem should cover your situation if you place suitable restrictions on your coefficient functions e.g. that they have at most linear growth. The paper also has references to alternative approaches such as the use of the Malliavin calculus.
