# Show that if the sequence $X_1, X_2,\ldots$ converges to a constant $\theta$ in probability then it converges to $\theta$ in distribution

Show that if the sequence $X_1, X_2,\ldots$ converges to a constant $\theta$ in probability then it converges to $\theta$ in distribution

Can't seem to figure it out.... Any assistance would be greatly appreciated!

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It's worth noting that the converse is also true. – Nate Eldredge Oct 31 '11 at 12:58

A characterization of convergence in distribution of $(X_n)$ to $\theta$ is the fact that $\mathrm E(h(X_n))\to h(\theta)$ for every continuous bounded function $h$. Once you start with this, the road is clear: decompose the difference $\mathrm E(h(X_n)-h(\theta))$ into a part where $X_n$ is close to $\theta$ and a part where $X_n$ is not.
More precisely, assume that $|h(x)|\leqslant M$ for every $x$ and choose a positive $\varepsilon$. Since $h$ is continuous at $\theta$, there exists $\delta$ such that $|x-\theta|\leqslant \delta$ implies $|h(x)-h(\theta)|\leqslant\varepsilon$, hence $$|\mathrm E(h(X_n))-h(\theta)|\leqslant \mathrm E(|h(X_n)-h(\theta)|)\leqslant2M\mathrm P(|X_n-\theta|\geqslant\delta)+\varepsilon.$$ When $n\to\infty$, $\mathrm P(|X_n-\theta|\geqslant\delta)\to0$ hence $\limsup|\mathrm E(h(X_n))-h(\theta)|\leqslant\varepsilon$. This holds for every positive $\varepsilon$ hence $\mathrm E(h(X_n))\to h(\theta)$.