uniform limit of a sequence of functions whose left-limit exists, also has a left limit let $(f_{n})_{n}$ be a sequence of functions that are in the space of functions with left limit and that has a uniform limit $f$. How can I show that $f$ also has a left limit?
My idea: 
Consider a point $x$ and some $y < x$: 
it follows that $f(y)=\lim\limits_{n \to \infty}f_{n}(y)$
and $\lim\limits_{y \to x^{-}}f(y)=\lim\limits_{y \to x^{-}}\lim\limits_{n \to \infty}f_{n}(y)$
If I am allowed to switch limits $\lim\limits_{y \to x^{-}}\lim\limits_{n \to \infty}f_{n}(y)=\lim\limits_{n \to \infty}\lim\limits_{y \to x^{-}}f_{n}(y)=\lim\limits_{n \to \infty} f_{n}(x_{-})$
Is this sufficient and correct? Am I allowed to switch limits in this case?
 A: Let $x\in \mathbb{R}$ and $(x_n)_{n\in\mathbb{N}}\subseteq (-\infty,x)$ with $x_n\to x$.
Then, for every $m,n$ and $k$, 
\begin{align}
|f(x_n)-f(x_m)| &\leq |f(x_n)-f_k(x_n)|+|f_k(x_n)-f_k(x_m)|+|f(x_m)-f_k(x_m)|\\
&\leq |f_k(x_n)-f_k(x_m)|+2\|f-f_k\|_{\infty}
\end{align}
Thus, given $\varepsilon>0$, we can fix $k$ such that $\|f-f_k\|_{\infty}\leq \frac{\varepsilon}{3}$ and thereafter $n,m$ large enough that $|f_k(x_n)-f_k(x_m)|<\frac{\varepsilon}{3}$. Accordingly, $(f(x_n))_{n\in \mathbb{N}}$ is Cauchy and hence, converges. 
Now, for uniqueness, if $(x_n)_{n\in\mathbb{N}}$ and $(y_n)_{n\in\mathbb{N}}$ are two sequences tending to $x$ from below, then define $z_n=x_{(n+1)/2}$ for $n$ odd and $z_n=y_{n/2}$ for $n$ even. Then, $z_n$ also satisfies the above assumptions, so $(f(z_n))_{n\in\mathbb{N}}$ is convergent. Since $(z_n)_{n\in\mathbb{N}}$ contains $(x_n)_{n\in\mathbb{N}}$ and $(y_n)_{n\in\mathbb{N}}$ as sub-sequences, this implies that 
$$
\lim_{n\to\infty} f(y_n)=\lim_{n\to\infty} f(z_n)=\lim_{n\to\infty} f(x_n)
$$
Accordingly, $\lim_{t\to x^-} f(t)$ exists, which is what we wanted.
A: Yes you are allowed to switch the order of limits here, and this is absolutely standard in the theory of uniform convergence.
Thm: Suppose $E\subset \mathbb R$ and  $x$ is a limit point of $E.$ Assume $f,f_1,f_2,\dots$ are real valued functions on $E$ and that $f_n\to f$ uniformly on $E.$ If for each $n,$
$$\lim_{y\to x, y\in E}f_n(y) = L_n\in \mathbb R,$$
then $\lim_{y\to x, y\in E}f(y) $ exists, and
$$\lim_{n\to \infty}\,\lim_{y\to x, y\in E}f_n(y) = \lim_{y\to x, y\in E}\,\lim_{n\to \infty}f_n(y) = \lim_{y\to x, y\in E}f(y).$$
This is Theorem 7.11 in baby Rudin, applied to $\mathbb R.$ In your problem we would take $E=(x-\delta,x)$ for some $\delta > 0.$ Within this $E,$ the approach to $x$ is from the left; there is no other choice. This gives the desired result for left limits.
