How to prove that if $f_n \rightarrow f$ in measure then $\dfrac{1}{f_n} \rightarrow \dfrac{1}{f}$? Let $f_n \rightarrow f$ in measure $\mu$ on A and $\mu(A) < +\infty$. If $f_n(x)$ and $f(x)$ are both not equal to 0 for all $x \in A$, then $\dfrac{1}{f_n} \rightarrow \dfrac{1}{f}$ in measure.
I don't know how to approach this question, as what I only know is the definition of function converging in measure. Plus, don't you think it's werid when $f_n(x)$ and $f(x)$ can also be $\infty$? I'm stuck at how to find my way around. Any hint for this would help a lot. 
 A: For a proof using the definition of convergence in measure, here's how you should think about this problem. You want to show that for each $r$
$$
\mu\left(\left|\frac1{f_n}-\frac1f\right|>r\right)=\mu\left(\frac{|f_n-f|}{|f_nf|}>r\right)
$$
tends to zero as $n\to\infty$. The problem is that the denominator $|f_nf|$ might be small, which  prevents the RHS from being small. Fortunately we can take advantage of the fact $\mu(f=0)=0$ to keep $f$ from getting too small, and since $f_n\to f$ in measure, this means that $|f_nf|$ can be kept away from zero "most of the time". Concretely, the fact
$$
0=\mu(|f|=0)=\mu\left(\bigcap_k \left\{|f|<\frac1k\right\}\right)=\lim_{k\to\infty}\mu\left(|f|<\frac1k\right)
$$
means that it will not cost us much to assume that $|f|\ge\delta$. Moreover, by convergence in measure it doesn't cost much to also assume $|f_n-f|\le\frac\delta2$; if so then $|f_n|\ge|f|-|f_n-f|\ge\frac\delta2$ and so $|f_nf|\ge\frac{\delta^2}2$. This reasoning leads to the following decomposition:
$$
\left\{\frac{|f_n-f|}{|f_nf|}>r\right\}\subset\left\{\frac{|f_n-f|}{|f_nf|}>r, |f|\ge\delta, |f_n-f|\le\frac\delta2\right\}\cup\left\{|f|<\delta \right\}\cup\left\{|f_n-f|>\frac\delta2\right\}
$$
where the comma stands for "and" (set intersection). As argued above, the first set on the RHS is a subset of $\left\{|f_n-f|>\frac {\delta^2}2r\right\}$ so 
$$
\begin{aligned}
\mu\left(\frac{|f_n-f|}{|f_nf|}>r\right)&\le\mu\left(|f_n-f|>\frac {\delta^2}2r\right)+\mu\left(|f|<\delta\right)+\mu\left(|f_n-f|>\frac\delta2\right)\\
&=A+B+C
\end{aligned}
$$
Now we see how to proceed. Given $r$ and $\epsilon$, choose $\delta$ so small that $B<\frac\epsilon2$. With this $\delta$ choose $N$ so large that $A+C<\frac\epsilon2$ whenever $n\ge N$.
A: This can be shown by a general theorem:
Theorem:
For any metric spaces $S$ and $T$, let $X_,X_1,X_2,\ldots$ random elements in $S$ and $f:S\rightarrow T$ continuous. If $X_n\xrightarrow{n\rightarrow\infty}X$ in probability, then $f(X_n)\xrightarrow{n\rightarrow\infty}f(X)$ ins probability
(See Kallenber's Foundation of Modern Probability, p.64)
which in turn follows from the following
Theorem:
$X_n$ converges in probability to $X$ iff every subsequence of $X_n$ has further subsequence that converges to $X$ a.s.
Proof:
For any $\varepsilon>0$
$$
\varepsilon\mathbb{1}_{\{d(X_n,X)>\varepsilon\}}\leq d(X_n,X)\wedge1\leq \varepsilon +\mathbb{1}_{\{d(X_n,X)>\varepsilon\}}
$$
Hence
$$
\varepsilon\mu\big(d(X_n,X)>\varepsilon\big)\leq\int d(X_n,X)\wedge1\,d\mu\leq \varepsilon\mu(\Omega)+\mu(d(X_n,X)>\varepsilon)
$$
Therefore, $X_n$ converges in measure to $X$ iff $D(X_n,X):=\int d(X_n,X)\wedge1\,d\mu\xrightarrow{n\rightarrow\infty}0$.
If $X_n\xrightarrow{n\rightarrow\infty}X$ in measure, then $D(X_n,X)\xrightarrow{n\rightarrow\infty}0$ and so, by standard result in integration, there is a subsequence $\{X_{n'}\}$ such that $d(X_{n'},X)\wedge1\xrightarrow{n'\rightarrow\infty}0$ $\mu$-a.s.
Conversely, suppose every subsequence of $\{X_{n'}\}$ of $\{X_n\}$ has a further subsequence $\{X_{n''}\}$ which converges to $X$ $\mu$-a.s. and yet, $X_n$ fails to converge to $X$ in measure. Then, there is $\varepsilon>0$ and  subsequence $\{X_{n_k}\}$ of $\{X_n\}$ such that 
$$
D(X_{n_k},X)>\varepsilon \tag{1}\label{one}
$$
 By assumption, there is a subsequence $\{X_{n_{k'}}\}$ of $\{X_{n_k}\}$ such that $X_{n_{k'}}$ converges to $X$ $\mu$-a.s. By dominated convergence $D(X_{n_{k'}},X)\xrightarrow{k''\rightarrow\infty}0$ which is a contradiction to $\eqref{one}$.

In the case at hand, $X,X_n$ are random elements in $S=\mathbb{R}\setminus\{0\}$ and $f:S\rightarrow\mathbb{R}$ given by $f(x)=1/x$ is continuous.
