The equivalence between almost surely continuous and continuous modification Consider the stochastic process $X:=(X_{t})_{t\in\mathbb{T}}$. We say $\tilde{X}$ is a modification of $X$ if for any fixed $t$, we have $\mathbb{P}\{\tilde{X}_{t}=X_{t}\}=1$.
Then, we have the Kolmogorov-Chentsov theorem that if $$\mathbb{E}|X_{t}-X_{s}|^{\alpha}\leq C|t-s|^{1+\beta},$$ for some $\alpha,\beta, C>0$, then there exists a continuous modification of $X$. 
Then my problem originates from the question:

why do we need $\beta>0$? Is $\mathbb{E}|X_{t}-X_{s}|\leq C|t-s|$ sufficient?

After some search, I found a counter-example arguing as follows:

Consider $\Omega:=[0,1]$ with its Borel $\sigma-$algebra and uniform probability measure. Let $U(\omega):=\omega$, that is $U$ is a Uniform$[0,1]$ random variable. Define $X_{t}(\omega):=\mathbb{1}_{U\leq t}(\omega),$ for $t\geq 0$. 
Then note that $|X_{t+h}-X_{t}|=1$ if $0\leq t<1$ and $t<U\leq t+h$, and $|X_{t+h}-X_{t}=0$ if otherwise. 
Thus, for all $0\leq t<1$ and $h>0$, we have $$\mathbb{E}|X_{t+h}-X_{t}|=\mathbb{P}(t<U\leq t+h)\leq h,$$ where the last inequality was obtained by considering the possibility of $t+h>1$. 
Thus, $\{X_{t}, t\geq 0\}$ satisfies the "insufficient inequality" with $C=1,\alpha=1,\beta=0$. 
However, consider $A:=\{\omega:t\mapsto X_{t}(\omega)\ \text{is continuous}\}$. Then if $\omega=0$, then $U(\omega)=0$ and $X_{t}(\omega)=1$ for all $t\geq 0$, so this particular sample path is continuous. If $\omega>0$, then $U(\omega)>0$ and clearly the associated sample path is not continuous. 
This implies that $A=\{0\}$, and thus $\mathbb{P}(A)=\mathbb{P}(\{0\})=0$.
Thus, $\{X_{t}, t\geq 0\}$ does not have continuous sample paths almost surely, which implies that it has not continuous modification.

I understand most of this argument, but cannot follows at the end of it. 
So it is basically saying that if a stochastic process is NOT a.s. continuous, then it cannot have continuous modificaiton.
Why is this true? Is almost surely continuous equivalent to having a continuous modification? or does one imply the other?
Thank you!
Edit 1 (One direction):
After some thought, I think the direction (having continuous modification) $\Rightarrow$ (almost surely continuous) is correct. I don't know if the inverse direction is correct, but $(\Rightarrow)$ is enough for me, since I showed that the process I defined is NOT almost surely continuous, and thus it cannot have continuous modification.
Let $\Omega_{0}\subset\Omega$ be with full measure, the definition of modification is equivalent to that:

the defined random variable $\tilde{X}:\Omega_{0}\times T\longrightarrow\mathbb{R}$ is called a modification of $X:\Omega\times\mathbb{T}\longrightarrow\mathbb{R}$, if $X$ and $\tilde{X}$ only differ on a set $\Omega\setminus\Omega_{0}$ of measure $0$.

This tells us, if $X_{t}$ has a modification $\tilde{X}_{t}$ that is continuous, then on $\Omega_{0}$, $X_{t}=\tilde{X}_{t}$ for all $t$, and thus $X_{t}$ is continuous on $\Omega_{0}$. 
This is exactly saying that $X_{t}$ is NOT continuous on at most on a null set, and thus it is at least almost surely continuous.
Edit 2:(Poisson Process)
As Saz pointed out, another example is Poisson process. 
For the Poisson process $\pi_{t}$ with parameter $\mu$, we have $$\mathbb{E}e^{i\lambda\pi_{t}}=e^{t\mu(e^{i\lambda-1})},$$ so that $\pi_{t}$ has Poisson distribution with parameter $t\mu$. In particular, $$\mathbb{E}|\pi_{t+h}-\pi_{t}|=\mathbb{E}\pi_{h}=h\mu,$$ so it satisfies the ``insufficient inequality'' with $C=\mu$. 
But the values of $\pi_{t}$ are integers and $\pi_{t}$ is not identically constant (the expectation grows), and thus $\pi_{t}$ does not have continuous sample paths almost surely (since you have integer value and you jump, so you measure the singleton set each time), and thus $\pi_{t}$ does not have continuous modification.
I think here we also used that if a process has continuous modification, then it is almost surely continuous.
By the way, I think $\Leftarrow$ is meaningless. If the process has been almost surely continuous, then there is no need for us to even consider modification, right? We need modification since almost sure continuity is rare in the stochastic process, isn't it?
So I think I asked a partially dumb question :(
As usual, I will leave the post open for a while and answer my own question to close it.
 A: Since the discussion stopped, I am gonna answer my own post to "close" it.
As I said in the edit, there seems no motivation to talk about the direction (a.s. continuity) $\Rightarrow$ (continuous modification). If it has been a.s. continuous, then there is no need to modify it. I have the modification only because the a.s. continuous is not really common in stochastic analysis.
(Direction $\Leftarrow$):
After some thought, I think the direction (having continuous modification) $\Rightarrow$ (almost surely continuous) is correct. I don't know if the inverse direction is correct, but $(\Rightarrow)$ is enough for me, since I showed that the process I defined is NOT almost surely continuous, and thus it cannot have continuous modification.
Let $\Omega_{0}\subset\Omega$ be with full measure, the definition of modification is equivalent to that:

The defined random variable $\tilde{X}:\Omega_{0}\times T\longrightarrow\mathbb{R}$ is called a modification of $X:\Omega\times\mathbb{T}\longrightarrow\mathbb{R}$, if $X$ and $\tilde{X}$ only differ on a set $\Omega\setminus\Omega_{0}$ of measure $0$.

This tells us, if $X_{t}$ has a modification $\tilde{X}_{t}$ that is continuous, then on $\Omega_{0}$, $X_{t}=\tilde{X}_{t}$ for all $t$, and thus $X_{t}$ is continuous on $\Omega_{0}$. 
This is exactly saying that $X_{t}$ is NOT continuous on at most on a null set, and thus it is at least almost surely continuous.
Another example:(Poisson Process)
As Saz pointed out, another example is Poisson process. 
For the Poisson process $\pi_{t}$ with parameter $\mu$, we have $$\mathbb{E}e^{i\lambda\pi_{t}}=e^{t\mu(e^{i\lambda-1})},$$ so that $\pi_{t}$ has Poisson distribution with parameter $t\mu$. In particular, $$\mathbb{E}|\pi_{t+h}-\pi_{t}|=\mathbb{E}\pi_{h}=h\mu,$$ so it satisfies the ``insufficient inequality'' with $C=\mu$. 
But the values of $\pi_{t}$ are integers and $\pi_{t}$ is not identically constant (the expectation grows), and thus $\pi_{t}$ does not have continuous sample paths almost surely (since you have integer value and you jump, so you measure the singleton set each time), and thus $\pi_{t}$ does not have continuous modification.
I think here we also used that if a process has continuous modification, then it is almost surely continuous.
