probability - For a non-negative absolutely continuous random variable $X$, with distribution $F$. Why is $lim_{trightarrow infty}t(1-F(t))=0$?

So I am given a non-negative absolutely continuous random variable $X$ with distribution $F$, and density $p_X$. I am given the definition of expectation using simple functions and the survival function. Using those I need to prove that $$E(X)=\int_0^\infty xp_X(x)dx$$

So I am given that $$E(X)=\int_0^\infty 1- F(t)dt$$

After applying integration by parts I get $$E(X)=\lim_{t\rightarrow \infty}t(1-F(t)) +\int_0^\infty xp_X(x)dx.$$

I am having trouble justifying why the first term is 0.

I have $$\lim_{t\rightarrow \infty}t(1-F(t))=\lim_{t\rightarrow\infty}t\int_t^\infty p_X(x)dx\leq\int_t^\infty xp_X(x)dx$$

But I dont know how to show the latter is 0.

Answer

Not an answer to the question in the title, but an alternative route:

Prescribe function $h$ by $\left(x,t\right)\mapsto1$ if $x>t$ and
$\left(x,t\right)\mapsto0$ otherwise. Then:

$$\int_{0}^{\infty}\left(1-F\left(t\right)\right)dt=\int_{0}^{\infty}\int_{0}^{\infty}h\left(x,t\right)p_{X}\left(x\right)dxdt=\int_{0}^{\infty}\int_{0}^{\infty}h\left(x,t\right)p_{X}\left(x\right)dtdx=$$$$\int_{0}^{\infty}p_{X}\left(x\right)\int_{0}^{\infty}h\left(x,t\right)dtdx=\int_{0}^{\infty}p_{X}\left(x\right)xdx$$

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edit2 (answer on question in title)

If $\mathbb E(X)<\infty$ then it can be shown that indeed $\lim_{t\rightarrow\infty}t\left(1-F\left(t\right)\right)=0$:

$$t\left(1-F\left(t\right)\right)=\int_{t}^{\infty}tp_{X}\left(x\right)dx\leq\int_{t}^{\infty}xp_{X}\left(x\right)dx$$
and $\int_{0}^{\infty}xp_{X}\left(x\right)dx=\mathbb E(X)<\infty$ tells us that $\lim_{t\rightarrow\infty}\int_{t}^{\infty}xp_{X}\left(x\right)dx=0$

If $\mathbb E(X)=\infty$ then you could take e.g. $F(t)=1-t^{-1}$ for $t>1$ as a counterexample.