I'm studying proof that if $W_t = (W_t^{(1)}, W_t^{(2)},..., W_t^{(d)})$ a $d$-dimensional Wiener process, $\mathcal{F}^W_t=\sigma(W_s: s\leq t)$ and $f:\mathbb{R}^d \rightarrow \mathbb{R}$ is a harmonic function such that $\mathbb{E}|f(W_t)| < \infty,$ then $(f(W_t), \mathcal{F}^W_t)$ is a martingale.
The proof proceeds with following equations which I have trouble to see why work - for $t>s:$ $$\mathbb{E}(f(W_t)|\mathcal{F}^W_s)=\mathbb{E}(f(W_s + (W_t - W_s))|\mathcal{F}^W_s)= (2\pi(t-s))^{-\frac{d}{2}}\int_{\mathbb{R}^d}f(W_s + x)e^{-\frac{|x|^2}{2(t-s)}}dx$$
I don't see how do we get $W_s$ under the integral.
By the Markov property, one has $\mathbb E[f(W_t)|\mathcal F_s^W]=\mathbb E[f(W_t)|W_s]$. One proceeds as in the question, using the fact that $W_t-W_s$ is independent of $W_s$. In general, if $X$ is $\mathcal G$-measurable, $Y$ is independent of $\mathcal G$ and $Y$ has distribution $\mu$, we have $$\mathbb E[f(X+Y)|X]=\int f(X+y)\mu(dy).$$