a question about a proof here involving real analysis or measure theory i have a question about a proof here that I was reading:

Basically what I don't understand is the last sentence of the proof where it says:
$Pr{\{t<S_{n+1} \leq t + \delta\}} = f_{S_{n+1}}(t)(\delta + o(\delta))$ is simply a consequence of the fact that ... has a continuous ... in the interval [t, t+$\delta$].
Could someone explains how to get from the left to the right of the equality by using the property of "continuous function" here?
Sorry I think it has to do with real-analysis, or measure theory? by my background is week, so I don't get how to use the "continuous function" property to arrive at what the author wrote here.
Thank you
 A: Very good question, I ran into this question for a different reason when I posted this question about the likelihood function for a Cox PH model;
Basic question regarding construction of likelihood function from a Cox PH model
The answer was related to the point you raise here about probabilities in arbitrarily small "elements" about a point (arbitrarily small intervals in $\mathbb{R}$ say) are equal to the density value at the point. This result comes from the fundamental relationship between $F(x)$ the distribution and $f(x)$ the density, that is $d/dxF(x)=f(x)$.
$F(x)$ here means the distribution function where $F(x)=\mu(-\infty,x]=P[X\leq x]$ and where $\mu$ is the probability law or distribution of $X$ satisfying the relationship
$$P[X\in A]=\mu(A)=\int_{A}f(x)dx$$
Thus putting $A=(-\infty,x]$ we get
$$\frac{d}{dx}P[X\in A]=\frac{d}{dx}P[X\leq x]
=\frac{d}{dx}\left[\int_{(-\infty,x]}f(x)dx\right]=f(x)$$
The underlying mathematics behind the $d/dx F(x)=f(x)$ is the fundamental theorem of calculus
https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
The link here
https://en.wikipedia.org/wiki/Likelihood_function#Likelihoods_for_continuous_distributions
given by the answer to my question shows the impact all this has on the likelihood function (i.e. maximising the likelihood for fixed small element about a point is the same as maximising the likelihood defined in terms of the density of that point).
In terms of your result shown from the paper, the link I gave shows the following limit applies;
$$\lim_{\epsilon\rightarrow 0^{+}}\frac{1}{\epsilon}\int_{(x,x+\epsilon]}f(x)dx=f(x)$$
and stems from this fundamental theorem of calculus, thus it is tempting (but not mathematically rigourous without the $o$ notation) to say
$$\int_{(x,x+\epsilon]}f(x)dx\approx\epsilon f(x)$$
and so from the previous relationship between $F$ and $f$
$$\int_{(x,x+\epsilon]}f(x)dx=\mu\left\{(x,x+\epsilon]\right\}=P[X\in(x,x+\epsilon]]\approx\epsilon f(x)$$
Thus without the formalitites of the $o$ notation I think of this result as saying the probability of $X$ being in a arbitrarily small element $(x,x+\epsilon]$ is the area of the arbitrarily narrow "rectangle" given by height=$f(x)$ multiplied by length $\epsilon$
EDIT: $o$-notation
For any function $g(x)$ with derivative $g'$ at $x_{0}$ by definition we have
\begin{align*}
\frac{g(x_{0}+\delta)-g(\delta)}{\delta}\underset{\delta\rightarrow 0}{\longrightarrow} g'(x_{0})
\end{align*}
Which implies
\begin{align*}
\frac{g(x_{0}+\delta)-g(\delta)-\delta g'(x_{0})}{\delta}\underset{\delta\rightarrow 0}{\longrightarrow} 0
\end{align*}
Let the notation $o(\delta)$ be a function of $\delta$ such that $o(\delta)/\delta\longrightarrow 0$ as $\delta\longrightarrow 0$. For example $C\delta^{2}=o(\delta)$ for any constant $C$ will do since $(C\delta^{2})/\delta=C\delta\longrightarrow 0$ as $\delta\longrightarrow 0$. Similarly $[C o(\delta)]/\delta=C[o(\delta)/\delta]\longrightarrow 0$ by definition of $o(\delta)$ so that $C o(\delta)$ is $o(\delta)$. But $C\delta$ is not $o(\delta)$ since $(C\delta)/\delta=C$ which is constant. The idea is that many functions could be $o(\delta)$ but the lack of uniqueness does not matter - only the limiting behaviour matters. With this in mind from above
\begin{align*}
\frac{g(x_{0}+\delta)-g(\delta)-\delta g'(x_{0})}{\delta}=o(\delta)=C o(\delta)
\end{align*}
which gives
\begin{align*}
g(x_{0}+\delta)-g(\delta)=\delta[ g'(x_{0})+ Co(\delta)]
\end{align*}
Replacing $g$ with $F$, $f'$ with $F'=f$, $x_{0}$ with $t_{0}$ and choosing $C=f(t_{0})$ we get
\begin{align*}
F(t_{0}+\delta)-F(\delta)&=\delta[ f(t_{0})+ f(t_{0})o(\delta)]\\
&=\delta f(t_{0})+\delta f(t_{0})o(\delta)\\
&=f(t_{0})[\delta+\delta o(\delta)]
\end{align*}
But $[\delta o(\delta)]/\delta=o(\delta)$ so that $\delta o(\delta)$ is itself $o(\delta)$. Thus
\begin{align*}
F(t_{0}+\delta)-F(\delta)&=f(t_{0})[\delta+ o(\delta)]
\end{align*}
Thus
\begin{align*}
Pr\{X\in (t_{0},t_{0}+\delta]\}=f(t_{0})[\delta+ o(\delta)]
\end{align*}
