Problem: Let $X$ be uniformly distributed in the unit interval $[0,1]$, let \begin{equation*} Y_n= \begin{cases} 0, & \text{if}\:X \geq \dfrac{1}{n} \\ \\ n, & \text{if}\:X < \dfrac{1}{n} \end{cases} \end{equation*}
a) Show the sequence $Y_n$ converges to $0$.
b) Prove $E[Y_n]=\dfrac{1}{2}$
$\underline{\text{My Attempt}}$
a) Convergence in probability concept can be used. But using the intution since $X \sim \text{Uni}[0,1]$ therefore
$P(X<\frac{1}{n})=\frac{1}{n}$ and $P(X \geq \frac{1}{n})=1-\frac{1}{n}$
Hence as $n \to \infty$
$P(Y_n = 0) = 1$ and $P(Y_n = n) = 0$
Therefor $Y_n$ converges to $0$.
b) \begin{align*} E[Y_n] &= E[E[Y_n|X]] \\ &=P(X<\frac{1}{n})E[Y_n|X<\frac{1}{n}]+P(X\geq\frac{1}{n})\underbrace{E[Y_n|X\geq\frac{1}{n}]}_{0}\\ &=\frac{1}{n}nP(Y_n|X<\frac{1}{n})+0\\ &=\frac{1}{n}n\dfrac{P(\{Y_n=n\} \cap \{X<\frac{1}{n}\})}{P(X<\frac{1}{n})}\\ &=\frac{1}{n}n\dfrac{\frac{1}{n}}{\frac{1}{n}}\\ &=\frac{1}{n}n\cdot 1\\ &=1 \end{align*}
I am thankful to the author of the book, they have confirmed that it was a mistake in the book so the summation is not required since it is a sequence therefore the answer is $1$ as one of the person comment (Thank you
Dominik Kutek).