(Dis)prove continuity, integrability of $f:[0,1] \rightarrow \mathbb{R}$ with 'jumps'.

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Let $f:[0,1] \rightarrow \mathbb{R}$ be

$f(x)= \begin{cases} x \text{ if } x \neq \frac{1}{n} \text{ for all } n \in \mathbb{N} \\ 1 \text{ if } x =\frac{1}{n} \text{ for some } n \in \mathbb{N} \end{cases}$

Now first I want to answer the question if $f \in C([0,1])$, i.e. is $f$ continuous on the interval.

For continuity I would like to use the following definition

$$\forall \epsilon>0, \exists \delta>0, \forall x,y \in [0,1] : |x-y|<\delta \Rightarrow |f(x)-f(y)| < \epsilon. $$

I argue the function is not continuous since it has 'jumps', but how to formally prove this? Intuitively I want to create a $\epsilon$ close to one of the jumps, for example close to $f(\frac{1}{2})$. I see that this would mean that any $y$ within $\delta$ of $x$ would result in a $|f(y)-f(x)|$ of at least $\epsilon-\delta$. Is this enough?

Then for integrability. I would like to use the definition

$f:[0,1] \rightarrow \mathbb{R}$ is integrabel if

$$\forall \epsilon>0, \exists P: 0 \leq \bar{S}_{P}-\underline{S}_{P} \leq \epsilon, $$ where $\bar{S}_{P}=\sum\limits_{j=1}^{N}M_{j}(x_{j}-x_{j-1})$ where $M_{j}$ is the supremum for each interval.

Now intuitively I would say the function is integrabel. I want to create very small intervals around the jumps, and thereby contain the effect these jumps (of value $f(x)=1$) have on the total sums. So the partition should have a form of something like:

Let the partition $P$ be an addition of two sums, where one sum contains all intervals with the jumps and the other one contains the rest. Let us contain all the $x$ values with a jump in $(\frac{1}{n}-\delta),(\frac{1}{n}+\delta)$. Would this approach lead to anything?

All help or suggestions are welcome!

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Your function is not continuous at $x=0$ because you a have the sequence $\{1/n\}$ approaching $0$ but the sequence $\{ f(1/n)\}$ does not approach $f(0)$.

Note that $f(1/n)=1$ while $f(0)=0$

Your function is integrable but your partition does not work because the same $\delta$ will add up and generates trouble in your Riemann's sum.

Try another partition which total measure (length) is less than $\epsilon$ and you will be on the right track.