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Consider the euclidean space $X = \mathbb{R}^d, \; d \geq 1$, and $\mathcal{B}(\mathbb{R}^d)$ the Borel $\sigma$-algebra. For every $x \in \mathbb{R}^d$, $A \in \mathcal{B}(\mathbb{R}^d)$ and $t > 0$ let define $$P_t(x, A) = \frac{1}{\sqrt{(2 \pi t)^d}} \int_A e^{- \frac{| x - y |^2}{2t}} dy.$$ Let $\mu$ be a probability measure on $X$.

Can someone give me a hint on how can I compute the following integral: $$T_t \mu(A) := \int_X P_t(x, A) d\mu(x) \; ?$$

P.S. Let $\mathcal{M}(X)$ be the space of all signed measures defined on $X$. Then $T_t : \mathcal{M}(X) \to \mathcal{M}(X)$.

Thank you!

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    What exactly do you mean by "compute the [...] integral"? – 2017-02-12
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    I want to find a simpler form for the operator $T_t$. In fact, I want to verify if the sequence $$\frac{1}{\tau_n} \int_0^{\tau_n} T_t \mu dt$$ is convergent in the weak topology, where $\tau_n$ is a sequence of positive numbers such that $\tau_n \nearrow \infty$. – 2017-02-12
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    You can change the order of integration. I don't think you can simplify it beyond that. – 2017-02-13

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Referring to your comment above, I will prove that $\frac{1}{t} \int_0^t T_s \mu \; ds$ does not converge weakly to any measure as $t \to \infty$.

First, changing variables with $u=t^{-1/2}(y-x)$ gives $$P_t(x,A) = (2\pi)^{-d/2} \int_{\Bbb R^d} e^{-|u|^2/2}1_{[t^{-1/2}A+x]}\;du$$ Since $1_{[t^{-1/2}A+x]} \to 0$ pointwise as $t \to \infty$ for any bounded borel set $A$, the dominated convergence theorem tells us that $P_t(x,A) \to 0$ as $t \to \infty$ for any bounded Borel set $A$. It follows that the Cesaro means $t^{-1}\int_0^t P_s(x,A)ds$ also converge to zero as $t \to \infty$.

By applying Fubini's theorem, we see that for any probability measure $\mu$ on $\Bbb R^d$ and any bounded Borel set $A$, $$t^{-1}\int_0^t (T_s\mu)(A)ds = \int_{\Bbb R^d} \bigg(t^{-1}\int_0^tP_t(x,A)ds \bigg)\mu(dx) \stackrel{t\to \infty}{\longrightarrow} 0$$ where the final limit is obtained by our preceding observations and the dominated convergence theorem (which is justified since the integrand is bounded above by $1$, which is $\mu$-integrable since $\mu$ is a probability). This proves the claim.

Remark: What this conclusion basically means is that there is no invariant probability measure for the heat semigroup (i.e, for Brownian motion) on $\Bbb R^d$. However, as you may easily check, there are certainly invariant measures which are non-probability, for instance Lebesgue measure.