I have a question: Consider two sequences of continuous, bijective functions $f\_n$ and $g\_n$ mapping $\mathbb{R}\to\mathbb{R}$. I know that for every compact $K\in\mathbb{R}$ that 


$$\lim\_{n\to\infty}\int\_K |f\_n(x)-g\_n(x)| \mathrm d x =0.$$


Furthermore the limit $f$ is surjective and continuous and


$$\lim\_{n\to\infty}\int\_K |f\_n(x)-f(x)| \mathrm d x =0,$$


Now, does this also hold for the inverse functions $f^{-1}\_n,g^{-1}\_n$, i.e.


$$\lim\_{n\to\infty}\int\_K |f\_n^{-1}(x)-g\_n^{-1}(x)| \mathrm d x =0?$$


If not, is there at least a weaker form of convergence? Thanks for the answer :)



Convergence of sequence of inverse functions
The statement is in fact true if we assume that $f\_n$ and $g\_n$ are bijective and continuous functions form $\mathbb{R}$ to $\mathbb{R}$, hence monotone, say increasing, and converge point-wise to a surjective function $f$ (hence monotone and continuous). 


Indeed let $K\subset [a,b]\subset\mathbb{R}$. Since $f$ is surjective, there are $A\in\mathbb{R}$ and $B\in\mathbb{R}$ such that $f(A)<a$ and $f(B)>b$ and by the point-wise convergence, there holds $f\_n(A)<a$, $g\_n(A)<a$, $f\_n(B)>b$, $g\_n(B)>b$ eventually as $n\to\infty$. Now 
$\int\_a^b|f\_n^{-1}(s)-g\_n^{-1}(s)|ds$ is the two-dimensional measure of the set
$$ \big\{(s,t)\in[a,b]\times\mathbb{R} : \ f\_n^{-1}(s)\wedge g\_n^{-1}(s)\le t\le f\_n^{-1}(s)\vee g\_n^{-1}(s)\big\}$$
which is eventually included in the set 
$$ \big\{(s,t)\in\mathbb{R} \times[A,B] : \ f\_n(t)\wedge g\_n(t)\le s\le f\_n(t)\vee g\_n(t)\big\},$$
whose Lebesgue measure is $\int\_A^B|f\_n(t)-g\_n(t)|dt$. Since by assumption the latter converges to $0$, so does $\int\_K|f\_n^{-1}(s)-g\_n^{-1}(s)|ds$.


$$\*$$


Also note that the function $f(x)=\lim f\_n(x)$ is increasing, and being surjective, it is also continuous. Also, in these hypotheses $f\_n$ and $g\_n$ converge uniformly on compact sets, in fact, it would sufficient to assume $f\_n\to f$ and $g\_n\to f$ in $L^1\_{\it loc}$.


