Discontinuous points of $0.a_{1}0a_{2}0...$ Let’s assume we have a function called $f$ with the domain $(0, 1)$, and let’s also assume that we replace any number like $a + 0.99999...$ with $a + 1$. 
If $f$ is defined like this:
$$f(0.a_{1}a_{2}a_{3}...) = 0.a_{1}0a_{2}0a_{3}0...$$
What are the discontinuous and continuous points of $f$? How can we prove the answer?
 A: Short, intuitive answer:
If $x$ has infinite decimal expansion, you can change it as far back in that expansion as you want, and the value of $f(x)$ will only change (twice as) far back in its expansion.
If $x$ has terminating decimal expansion, then even the smallest decrease in $x$ must change the expansion of $x$ at the terminating digit (or earlier). This turns your to give a jump in the value of $f$.
A: This is more of a comment/addendum to Arthur's answer, which is however much to long for a comment. I think the argument in case of the non-terminating decimal expansion needs to be exapnded a bit. 
The thing is, if you have say $x=0.50\ldots0\overline{1234}$, and there are $1000$ zero digits between the initial $5$ and the periodical part, then decreasing the number by $\Delta x=10^{-1001}$ will give $x-\Delta x=0.49\ldots9\overline{1234}$ and $f(x-\Delta x)=0.409\ldots$, that means the very small change by $\Delta x$ caused a very big change in $f$. 
The argument goes that for a non-terminating expansion of $x$, for each integer $n$ you can find an integer $N > n$ such the the $N$-th decimal digit of $x$ $d_N$ is neither 
$0$ nor $9$. That means if you change $x$ by $\Delta x$ with $\lvert\Delta x\rvert < 10^{-N}$, you are changing $d_N$ by at most $1$ (in either direction), and since $d_N \neq 0,9$, that cannot cause an overcarry into the previuous digit (if $x$ was increased) or an undercarry into the previous digit (if $x$ was decreased). 
So in either cases, the decimal expansion of $x+\Delta x$ will be the same as for $x$ up to at least the $(N-1)$-th decimal, so their $f$ values will be the same at least until the $(2N-2)$-th decimal digit, as Arthur said.
A: Consider
$$f_n(x) = \frac{\{10^{n}x\}}{10^{2n}}-\frac{\{10^{n+1}x\}}{10^{2n+1}},\ \ n=0,1,2\dots,$$
where $\{ x\}$ is fractional part function. Note that $f_n(x)$ is right-continuous everywhere and continuous at all non terminating (rational or irrational) decimals.
Now, your $f(x)$ is just
$$f(x)=\sum_{n=0}^{\infty}f_n(x).$$
The series converges unformly, by Weierstrass M-test. Since uniform convergence preserves continuity, we conclude that $f(x)$ too is right-continuous everywhere and continuous at all non terminating decimals.
Generalization of Arthur's example in comments leads to the conclusion that the function is not left-continuous at terminating decimals numbers.
You might want to have a look at this similar function, where all continuities and discontinuities are dealt with using decimal expansions.
