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Let $X$ be a non-negative random variable, and suppose that $P(X \geq n) \geq 1/n$ for each $n \in \mathbb{N}$. Prove that $E(X) = \infty$.

I have been stuck with this problem for a few days now. I guess it can make some sense intuitively because you have some probability mass everywhere, and we're looking at probability of it being greater than some value. I tried to use inequalities like Markov's and Chebyshev's with no luck. I was hoping if someone can please explain to me how to answer this problem. It is coming from an introductory probability with measure theory book, and I am trying my best to get better at these kind of problems.

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Suppose we have a discrete random variable $Y$ with $\mathbb P(Y=n) = \frac{1}{n(n+1)}$ for all positive integers $n$

then $\mathbb P(Y\ge n) = \frac1n \le \mathbb P(X \ge n)$ for all positive integers $n$

and $\mathbb P(Y\ge x) \le \mathbb P(X \ge x)$ for all real $x$

so $X$ has weak first-order stochastic dominance over $Y$

making $\mathbb E[Y]\le \mathbb E[X]$

but $\mathbb E[Y] = \sum\limits_{n=1}^\infty \frac{1}{n+1}= +\infty$

showing $\mathbb E[X] = +\infty$

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I am giving a proof for the continuous case. Integrating by parts in the definition of expected value and observing that $P(X<-t)=0$ for any $t>0$ you have $$ \mathbb{E}[X]=\int\limits_0^{\infty}P(X>t)\,dt\ge \int\limits_0^{\infty}\frac{1}{t+1}\,dt=\infty $$ (since $P(x>t)\ge P(x>{\rm{ceiling\ of}}\, t)$ and ${\rm{ceiling\ of}}\, t\le t+1$

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