Suppose $X_1,X_2$ are iid random variables with common N(0,1) distribution. Define $$Y_n = \frac{X_1}{\frac{1}{n}+|X_2|}$$. Use Fubini's Theorem to verify that $$E(Y_n)=0$$ Note that as as $n \rightarrow \infty$ $$Y_n \rightarrow Y:= \frac{X_1}{|X_2|}$$ and that the expectation of $Y$ does not exists.
Proof:
$E(Y_n)= \int_{\Omega_1 \times \Omega_2} Y_n dP=\int_{\Omega_1}\big(\int_{\Omega_2} \frac{X_{1}(\omega_1)}{\frac{1}{n}+|X_{2}(\omega_2)|} dP_{2}(\omega_2)\big)dP_1(\omega_1)=\int_{\Omega_1}X_{1}(\omega_1)\big(\int_{\Omega_2} \frac{1}{\frac{1}{n}+|X_{2}(\omega_2)|} dP_{2}(\omega_2)\big)dP_1(\omega_1)=\int_{\Omega_2} \frac{1}{\frac{1}{n}+|X_{2}(\omega_2)|} \big(\int_{\Omega_1}X_{1}(\omega_1)\big(dP_1(\omega_1) \big)dP_{2}(\omega_2)=\int_{\Omega_2} \frac{1}{\frac{1}{n}+|X_{2}(\omega_2)|} E(X_1)dP_{2}(\omega_2)=\int_{\Omega_2} 0 *\frac{1}{\frac{1}{n}+|X_{2}(\omega_2)|} dP_{2}(\omega_2)=\int_{\Omega_2} 0 dP_{2}(\omega_2)=0$
You're nearly there. Try putting the integral over $\Omega_1$ on the inside:
\begin{align} \mathbb{E}(Y_n) &= \int\limits_{\Omega_1\times \Omega_2}Y_n dP = \int\limits_{\Omega_2} \int\limits_{\Omega_1} \frac{X_1(\omega_1)}{\frac{1}{n} + |X_2(\omega_2)|} dP_1(\omega_1) dP_2(\omega_2) \\ &= \int\limits_{\Omega_2} \frac{1}{\frac{1}{n} + |X_2(\omega_2)|} \left(\int\limits_{\Omega_1} X_1(\omega_1) dP_1(\omega_1) \right)dP_2(\omega_2) \\ &=\int\limits_{\Omega_2} \frac{1}{\frac{1}{n} + |X_2(\omega_2)|} \left(0\right)dP_2(\omega_2) \\ &= 0. \end{align}