Assume $(X_i)_{i=1,...,n}$ are a sequence of real iid random variables with continuous density $p_x$.
We know that $$Y:=\sum_{i=1}^n 1\{X_i\leq u\}\sim Bin(n,F_x(u)),$$ since $1\{X_i\leq u\}\sim Ber(F_x(u))$ with $F_x(u):=P(X_i\leq u)$.
By $X_{[i]}$ we denote the ith order statistic. We know that $$X_{[j]}\leq u\Leftrightarrow X_{[i]}\leq u ~(\forall i\leq j) \Leftrightarrow \sum_{i=1}^n 1\{X_i\leq u\} \geq j.$$
Consequently, it holds that $$P(X_{[j]}\leq u)=P(Y\geq j)=\sum_{k=j}^n \binom{n}{k}F_x^k(u)\left(1-F_x(u)\right)^{n-k}.$$
My question is the following: Why do I have to assume that $p_x$ is continuous? Don't I have the same binomial distribution if the $X_i$s are discrete random variables?
My professor stated that the problem is that for discrete random variables there could occur ties, i.e., $P(X_i=X_j)\neq 0$ for $i\neq j$, making the proof above false; but I don't see the problem.