Let $f$ be a nonegative integrable function.
Show that the function $F$ defined by $F(x)=\int_{-\infty}^{x}f$
is continuous by using Monotone Convergence Theorem.
The solution by Royden is:
Let $f_n=f\mathcal{X}_{(-\infty,x-1/n]}$, $(f_n)$ is an increasing sequence of nonnegative measurable functions with $\lim_n f_n=f\mathcal{X}_{(-\infty,x]}$. By MCT, $\lim_n F(x-1/n)=\lim_n \int f_n=\int f\mathcal{X}_{(-\infty,x]}=F(x)$.\
Similary, with $g_n=f\mathcal{X}_{(x+1/n,\infty)}, \lim_n F(x+1/n)=F(x)$.
Now, given $\epsilon>0$, there exists $N$ such that $F(x)-F(x-1/n)<\epsilon$ and $F(x+1/n)-F(x)<\epsilon$ whenever $n\geq N$.
All the above I understand.
What I do not understand is the following:
Now, choose $\delta<1/N$, then $|F(x)-F(y)|<\epsilon$ whenever $|x-y|<\delta.$
Hence $F$ is continuous.
Why?
I know that $F$ is continuous in $x$ if and only if FOR ALL $(x_n)$ sequence with $x_n \to x$ then $F(x_n)\to F(x)$.
But in our case, we only tested it for the sequence $x +1/n$ and $x-1/n$
Why is it enough to prove continuity with these two successions?
Best Answer
Function $F(x)$ is monotonically increasing, because $f$ is nonnegative.
Combining this with $\lim_{n\to\infty}F\left(x-\frac1n\right)=F(x)=\lim_{n\to\infty}F\left(x-\frac1n\right)$ it can be proved that $\lim_{n\to\infty}F(x_n)=x$ is true for every sequence $(x_n)_n$ converging to $x$.
Let $(x_n)_n$ be a convergent sequence in $\mathbb R$ and let $x$ denote its limit.
WLOG we may assume that $x-1<x_n<x+1$ for every $n$.
For every $n\in\mathbb N$ let $k_n$ be the largest positive integer with:$$x-\frac1{k_n}<x_n<x+\frac1{k_n}$$
Then evidently $\lim_{n\to\infty}k_n=\infty$ and: $$F\left(x-\frac1{k_n}\right)\leq F(x_n)\leq F\left(x+\frac1{k_n}\right)$$
From $\lim_{n\to\infty}k_n=\infty$ it follows that $F\left(x-\frac1{k_n}\right)\to F(x)$ and $F\left(x+\frac1{k_n}\right)\to F(x)$ (as shown in your question).
This allows the conclusion that $F(x_n)\to F(x)$.