This is a homework question. I think I have the correct answer, but I am not sure. Also, the wording sounds very awkward. Is there a better way to show this (or better way to word this)?
Let $X_1,\ldots,X_n$ be a random sample from $f_\theta(x)$, where $f_\theta(x)$ is a continuous density with support $[\theta,\infty)$. Show that the minimum converges in distribution to $\theta$, i.e. $X_{(1)}\overset{P}{\longrightarrow}\theta$.
MY ATTEMPT:
The distribution function is $F(x)=\int_\theta^x f_\theta(t)dt$. Now we find the distribution function of the minimum\begin{align*} 1-F_{X_{(1)}}(x) &= P(X_{(1)} > x) \\ &= P(X_1>x,\ldots, X_n>x)\\ &= P(X_1>x)^n\\ &= (1-F(x))^n\\ F_{X_{(1)}}(x) &= 1-\left(1-\int_\theta^x f_\theta(t)dt\right)^n \end{align*} For $x< \theta, F(x)=0$ so therefore $F_{X_{(1)}}(x) = 1-(1-0)^n=0$ for all $n$. For $x\geq \theta$ suppose that $F(x) = c$ where $0\leq c \leq 1$. Now we have $$\lim_{n\to\infty} F_{X_{(1)}}(x) =\lim_{n\to\infty} 1-(1-F(x))^n = \lim_{n\to\infty} 1-(1-c)^n = 1$$ Therefore $$ F_{X_{(1)}}(x) \longrightarrow G(x) = \left\{ \begin{array}{l l} 1 & \quad \text{if $x\geq \theta$}\\ 0 & \quad \text{if $x < \theta$} \end{array} \right.$$ This is a distribution function which puts all probability on one point, $\theta$. Therefore $X_{(1)}\overset{D}{\longrightarrow} \theta$, and since $\theta$ is a constant, $X_{(1)}\overset{P}{\longrightarrow}\theta$.
