Showing that $E(X\log X)\le 1$ Let $\{X_n\}, X\in L^1(\Omega, \mathcal{F},P)$ which satisfy the following properties:
(i) For all $n$, $X_n\ge 0, \ E(X_n)=1$ and $E(X_n\log X_n)\le 1$.
(ii) For any bounded random variable $Y$, $E(X_n Y)\to E(XY)$ as $n\to\infty$ 
Show that $X\ge 0$ with probability $1$ and $E(X\log X)\le 1$.
Note: I could prove that $X\ge 0$ with probability $1$. Also, from non-negativity of $X_n$ and $X$, we know that $X_n$ are $L^1 -$ bounded, hence they are tight. Does this help to prove that $E(X \log X)\le 1$? 
 A: Here I came up with a solution, hoping that it is correct. Let us prove the result in the following steps:
Step 1:
For any random variables $X,Y\ge 0$ with $E(Y)\le E(X)$, we have $E(X \log X)\ge E(X\log Y)$.
Proof:
Using the inequality $\log x\le x-1$ and putting $x=\frac{Y}{X}$, we get : $X(\log Y-\log X)\le Y-X\implies E(X\log Y- X\log X)\le E(Y)-E(X)\le 0$. Hence the proof follows.
Step 2 :
Define $Y_n=1$ if $X\in [\frac{1}{n},n]^c$ and $Y_n=X$ if $X\in [\frac{1}{n},n]$. Clearly, $\log Y_n$ is a bounded random variable. Now, $E(Y_n)=E(X\mathbb{I}\{X\in [\frac{1}{n},n]\})+P(X<\frac{1}{n})+P(X>n)$. Since, $E(X\log X)$ exists, $P(X=0)=0$. Thus, $E(Y_n)\to 1$ as $n\to\infty$. Fix any $\alpha>1$. Then, there exists $N$ such that for all $n\ge N$, $E(Y_n)<\alpha$. Now, using step 1, since $E(\alpha X_m)=\alpha$ for all $m\ge 1$, we can write:
$$E((\alpha X_m)\log (\alpha X_m))\ge E(\alpha X_m\log Y_n) \ \ \forall m\ge 1, n>N \\
\implies \alpha\log \alpha E(X_m)+\alpha E(X_m\log X_m)\ge \alpha E(X_m\log Y_n) \ \ \forall m\ge 1, n>N \\
\implies \alpha\log\alpha+\alpha\ge\alpha E(X_m\log Y_n) \ \ \forall m\ge 1, n>N.$$ Now, using boundedness of $Y_n$, letting $m\to\infty$, we get that $$\alpha\log\alpha+\alpha\ge\alpha E(X\log Y_n)=\alpha E(X\log X\ \mathbb{I}\{X\in [1/n,n]\}) \ \ \forall n>N.$$ Now, letting $n\to\infty$, we get 
$$\alpha\log\alpha+\alpha\ge\alpha E(X\log X).$$ Since $\alpha>1$ is arbitrary, letting $\alpha\downarrow 1$, we finally get $E(X\log X)\le 1$.
