Let $\xi:=(\xi_{n})_{n\geq 0}$ be IID and $\eta:=(\eta_{n})_{n \geq 0}$:
A path is defined as a map for fixed $\omega$ that $[0,\infty[\ni t\mapsto\xi_{t}(\omega)$
Show that:
$\xi$ has modification $\eta$ with continuous paths $\iff \exists c \in \mathbb R$ s.t. $P(\xi_{0}=c)=1$
I assume that the $\Rightarrow$ is easiest to prove and I guess it is easiest via contradiction.
So assume that for any $c \in \mathbb R$ that $P(\xi_{0}=c)<1$ then there must exist some set $A$ where $\{\xi_{0}=c\}\cap A=\varnothing$ and $P(A)>0$, but since $\xi$ has modification $\eta$ with continuous paths, then $P(\eta_{0}=c)<1$ and thus $\{\eta_{0}=c\}\cap A=\varnothing$. Now we need to use the continuity of our map, but I am not sure whether I am even on the right path. Any hints?
No generality is lost by assuming that all of the random variables $\xi_n$, $\eta_n$ are bounded by one fixed constant. If not, replace them by $\arctan\xi_n$ and $\arctan\eta_n$ and proceed.
The continuity of $\eta$ implies that $(\omega,n)\to\eta_n(\omega)$ is jointly measurable, permitting the use of Fubini's theorem. Let $c=\Bbb E[\eta_n]=\Bbb E[\xi_n]$. Use Fubini to show that $$ \Bbb E\left[\left(\int_a^b\eta_t\,dt\right)^2\right]=(b-a)^2c^2, $$ for each choice of $0\le a<b<\infty$. From this you see that the variance of the random variable $\int_a^b\eta_t\,dt$ is zero. Therefore $\int_a^b\eta_t\,dt = (b-a)c$, for all $0\le a<b<\infty$, a.s. From this and the continuity of $\eta$ it follows that $\Bbb P[\eta_t=c$ for all $t\ge 0]=1$. In particular, $\Bbb P[\xi_0=c]=\Bbb P[\eta_0=c]=1$.