Proof of posterior consistency Given the model 
$x_1,...,x_n$ ~ $Poi(\theta)$ (iid),
and the prior
$\theta$~$ Gamma(a,b)$.
How do we prove the consistency of posterior distribution?
Thanks in advance.
Modification:
The definition of posterior consistency given on my lecture slides is:
“given the data $x_1,...,x_n$ iid and the model f(x;$\theta$), if we have posterior consistency at $\theta_0$, then
$\forall \epsilon$,
   $\lim_{n\rightarrow +\infty}\pi(|\theta-\theta_0|>\epsilon|X)$=0.”
$\theta_0$ is not specified but I think it is the true value of $\theta$.
 A: Writing $X^{(n)}=(X_1,\dots,X_n)$, Bayes theorem tells us that
$$
  \Theta\mid X^{(n)}\sim\mathrm{Ga}\!\left(a+\sum_{i=1}^n X_i,\,b+n\right).
$$
Define
$$
  M=\mathrm{E}[\Theta\mid X^{(n)}]=\frac{a+\sum_{i=1}^n X_i}{b+n}
$$
and
$$
  V=\mathrm{Var}[\Theta\mid X^{(n)}]=\frac{a+\sum_{i=1}^n X_i}{(b+n)^2}.
$$
For any given $\epsilon>0$, if $|\Theta-M|<\epsilon/2$ and $|M-\theta_0|<\epsilon/2$, then, using the triangle inequality, we have
$$
  |\Theta-\theta_0| = |\Theta-M+M-\theta_0|\leq |\Theta-M|+|M-\theta_0| < \epsilon.
$$
Considering the contrapositive of the previous statement, we see that
$$
  \{|\Theta-\theta_0|\geq\epsilon\} \subset \{|\Theta-M|\geq\epsilon/2\} \cup \{|M-\theta_0|\geq\epsilon/2\}.
$$
Hence, monotonicity and Chebyshev's (Чебышёв) inequality give us
$$
  \Pr(|\Theta-\theta_0|\geq\epsilon\mid X^{(n)}) \leq \frac{4V}{\epsilon^2} + \Pr(|M-\theta_0|\geq\epsilon/2\mid X^{(n)}). 
$$
Let $\mu_0$ be a $\mathrm{Poi}(\theta_0)$ distribution and note that $V\to 0$ a.s. $[\mu_0]$, and $M\to \theta_0$ a.s. $[\mu_0]$, as $n\to\infty$.
This is essentially what you need to see that this posterior is "simulation consistent". That is, if you're simulating iid $X_i$'s from a $\mathrm{Poi}(\theta_0)$ distribution, then your posterior distribution will concentrate all its mass at $\theta_0$, as your sample size goes to infinity.
To learn how to write a precise proof, please study Schervish (Theory of Statistics), section 7.4.1 (Consistency of Posterior Distributions).
