Pointwise limit of the indicator function: $\lim_{c \to \infty}1_{\{X<kc\}}$ I want to find the pointwise limit of this indicator function:

$$\lim_{c \to \infty}1_{\{X&lt;kc\}}$$ 
where $k&gt;0$ is a parameter and $1$ is an indicator function and $X$ is finite random variable. 



One of my friends says it would be $\lim_{c \to \infty}1_{\{X&lt;kc\}}=1_{\{x&lt;\infty\}}=1$. Please provide me correct answer with explanation. Thank you.
Given $t$ in the domain of $X$, take $M$ so that $M&gt;\frac{X(t)}{k}$. This is possible because $X$ is finite and $k$ is fixed. Then, for all $c\geq M$, we have
$1_{X&lt;kc}(t) = 1$ because $X(t)&lt;Mk\leq ck$. Therefore $\lim_{c\to\infty}1_{X&lt;kc}(t)=1$ for each $t$, which means the pointwise limit $\lim_{c\to\infty}1_{\{X&lt;kc\}}$ is $1$.
$$\lim 1_{A_n} = 1_{\lim A_n}$$

So, convince yourself that $$\lim_{c \to \infty} \{X &lt; kc\} = \{X &lt; \infty\}$$ for $k&gt;0$