Let $X$ be a random variable such that $P(X>t) = \dfrac{e}{t\log t}$ for $t\geq e.$ If $Y_n = X1_{X\leq n}$, then what is wrong with the following calculation?
For $e\leq t\leq n,$ $P(Y_n>t) = P(Y_n>t|X\leq n)P(X\leq n)+P(Y_n>t|X>n)P(X>n) = \\ = P(X1_{X\leq n}>t|X\leq n)P(X\leq n)+0=P(X>t)P(X\leq n) = \\ =\dfrac{e}{t\log t}\Big(1-\dfrac{e}{n\log n}\Big).$
This is a problem from Durrett's Probability-theory and Examples and the correct answer is supposed to be simply $$P(Y_n>t) = \dfrac{e}{t\log t}.$$
Where exactly is the mistake in the above calculation?
By Bayes' conditional probability $$P(X1_{X\leq n}>t|X\leq n)= \frac{P((X1_{X\leq n}>t)\cap(X\leq n))}{P(X\leq n)} \\ =\frac{P(X>t)}{P(X\leq n)}$$ So your last line becomes $$P(Y_n>t)=\frac{P(X>t)}{P(X\leq n)}P(X\leq n)$$ which gives us the desired result