can you help me with the following inequality:
Let $T_x$ be the remaining life time of a $x$-year old Person and $\hat{e}_x = E[T_x]$. Then
$$\hat{e}_x \leq \hat{e}_{x+1}+1.$$
I know, that $\hat{e}_x = E[T_x] = \int_0^{\infty} t f_X(t) dt = \int_0^{\infty} \int_0^t 1 ds f_x(t)dt = \int_0^{\infty} \int_s^{\infty} f_X(t) dt ds = \int_0^{\infty} P(T_x > s) ds = \int_0^{\infty} p_{s,x} ds$
where $p_{s,x}$ is the probabilty of a $x-$ year old Person to survive $s$ years. Can I use a similar trick as above to Show the inequality?
If $T_x(\omega)$ denotes the remaining lifetime of a specific person having age $x$ for a specific scenario $\omega\in\Omega$, then $T_{x+1}(\omega)+1=T_{x}(\omega)$, since if he has age $x$ and dies at age $x+T_{x}(\omega)$, then the age of his death is $x+1+T_{x+1}(\omega)=x+T_{x}(\omega)$ and this yields $T_{x+1}(\omega)+1=T_x(\omega)$.
Now we have $$E[T_{x}]-1=E[T_x]=\int_0^\infty P(T_x>s)ds-1=\int_0^\infty P(T_x>s)-1_{[0,1]}(s) ds=\int_0^\infty E[1_{\{T_x>s\}}-1_{[0,1]}(s)] ds=\int_\Omega\int_0^\infty 1_{\{T_x(\omega)>s\}}-1_{[0,1]}(s) ds\ dP(\omega)\leq\int_\Omega\int_0^\infty 1_{\{T_{x+1}(\omega)>s\}} ds\ dP(\omega)=\ldots=E[T_{x+1}]$$