Show that $N_{t}:=\sup\{n:S_{n}\leq t\}$ is a stochastic process with time $\mathbb{R}_{\geq 0}$.

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I am working on an exercise about stochastic process as follows:

Let $(X_{n})_{n\in\mathbb{N}}$ be an i.i.d positive sequence. Define $S_{n}:=X_{1}+\cdots+X_{n}$, and $S_{0}=0$. Further, define $N_{t}:=\sup\{n:S_{n}\leq t\}$ for $t\geq 0$. Prove that $S_{t}:=S_{[t]}$ and $(N_{t})_{t\in\mathbb{R}_{\geq 0}}$ are stochastic process.

I have showed that $S_{t}$ is a stochastic process with time $\mathbb{T}:=\mathbb{R}_{\geq 0}$ by showing $S_{n}$ is a random variable for all $n\in \mathbb{Z}_{\geq 0}$. This is basically showing that if $X_{1}\cdots X_{n}$ are random variables, then so is $S_{n}$. In the case of $n=0$, we just have a constant which is a random variable.

I have a proof for $(N_{t})_{t\in\mathbb{R}_{\geq 0}}$ being a stochastic process, but I don't know if it is right.

Let's denote the probability space as $(\Omega,\mathcal{F},\mathbb{P})$.

Below is the proof:

To show $(N_{t})_{t\in\mathbb{R}\geq 0}$ is a stochastic process, we only need to show that $N_{t}$ is a random variable for all $t\in \mathbb{R}_{\geq 0}$.

But this follows from $$\{N_{t}< n\}=\{S_{n}>t\}\in\mathcal{F},$$ since we have showed that $S_{n}$ is a random variable for each $n\in\mathbb{Z}_{\geq 0}$.

I have doubt in my proof because:

It seems that showing $\{N_{t}< n\}\in\mathcal{F}$ for all non-negative integers $n$ is not sufficient for $N_{t}$ being a random variable. I think I need to show $\{N_{t}\leq r\}\in\mathcal{F}$ for all $r\in\mathbb{R}$. It seems that I only showed $N_{t}$ is a stopping time (if it is random)

What should I do? Thank you!

Edit 1: (Complete Proof)

Okay I think I figured it out. Here is my proof:

To show $(N_{t})_{t\in\mathbb{R}_{+}}$ is a stochastic process, fixing $t\in\mathbb{R}_{+}$, we need to show that $N_{t}$ is a random variable. To do this, let $x\in\mathbb{R}$, and consider $\{\omega:N_{t}(\omega)> x\}$.

Firstly, as $N_{t}(\omega)\in\mathbb{Z}_{\geq 0}$ for all $\omega\in\Omega$, we must have $\{\omega:N_{t}(\omega)>x\}=\Omega\in\mathcal{F}$ for all $x<0$. On the other hand, if $x\geq 0$, we have $$\{\omega:N_{t}(\omega)>x\}=\{\omega:N_{t}(\omega)\geq[x]+1\}=\{\omega:S_{[x]+1}(\omega)\leq t\}\in\mathcal{F},$$ the last set is in $\mathcal{F}$ because we have showed that $S_{n}$ is a random variable for all $n\in\mathbb{Z}_{\geq 0}$.

Thus, $\{\omega: N_{t}(\omega)>x\}\in\mathcal{F}$ for all $x\in\mathbb{R}$, and thus $N_{t}$ is a random variable. Since $t\in\mathbb{R}_{+}$ is arbitrarily fixed, we are done.

I will leave the post open for a few days in case I made a mistake or some further discussion happens. Then I will just answer my own post.

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Okay I think I figured it out. Here is my proof:

To show $(N_{t})_{t\in\mathbb{R}_{+}}$ is a stochastic process, fixing $t\in\mathbb{R}_{+}$, we need to show that $N_{t}$ is a random variable. To do this, let $x\in\mathbb{R}$, and consider $\{\omega:N_{t}(\omega)> x\}$.

Firstly, as $N_{t}(\omega)\in\mathbb{Z}_{\geq 0}$ for all $\omega\in\Omega$, we must have $\{\omega:N_{t}(\omega)>x\}=\Omega\in\mathcal{F}$ for all $x<0$. On the other hand, if $x\geq 0$, we have $$\{\omega:N_{t}(\omega)>x\}=\{\omega:N_{t}(\omega)\geq[x]+1\}=\{\omega:S_{[x]+1}(\omega)\leq t\}\in\mathcal{F},$$ the last set is in $\mathcal{F}$ because we have showed that $S_{n}$ is a random variable for all $n\in\mathbb{Z}_{\geq 0}$.

Thus, $\{\omega: N_{t}(\omega)>x\}\in\mathcal{F}$ for all $x\in\mathbb{R}$, and thus $N_{t}$ is a random variable. Since $t\in\mathbb{R}_{+}$ is arbitrarily fixed, we are done.