Prove the equality: $\sup\{x\geq0 : F_X(x)\leq t\}=\int_0^{+\infty} 1_{\{ F_X(x)\leq t\}}\,dx$.

Question

I was trying (unsuccessfully) to prove the following equality $$\sup\{x\geq0 : F_X(x)\leq t\}=\int_0^{+\infty} 1_{\{ F_X(x)\leq t\}}\,dx.$$ Can anyone give me an hint?

Answer 1

If $t\geq 1$, then the result is obvious (since both side are infinite). Suppose $t\in [0,1)$. Since $F_X$ is continuous and s.t. $$\lim_{x\to \infty }F_X(x)=1,$$ the set $\{x\geq 0\mid F_X(x)\leq t\},$ is non-empty and bounded. Set $$x_t=\sup\{x\geq 0\mid F_X(x)\leq t\}.$$

Since $F_X$ is non-decreasing and continuous, $$\{x\geq 0\mid F_X(x)\leq t\}=[0,x_t].$$

Therefore, $$\int_0^\infty \boldsymbol 1_{\{ F_X(x)\leq t\}}\,\mathrm d x=\int_0^\infty \boldsymbol 1_{[0,x_t]}(x)\,\mathrm d x=\int_0^{x_t}\,\mathrm d x=x_t=\sup\{x\geq 0\mid F_X(x)\leq t\},$$ as wished.