$\left\|f\right\|_{L^1(μ_1)}<∞$ $μ_2$-a.e.,$\left\|f\right\|_{L^1(μ_2)}<∞$ $μ_1$-a.e. $⇒$ $\left\|f\right\|_{L^1(μ_1\otimesμ_2)}<∞$

Question

Let $(\Omega_i,\mathcal A_i,\mu_i)$ be a $\sigma$-finite measure space and $f:\Omega_1\times\Omega_2\to\mathbb R$ be measurable with respect to $\mathcal A_1\otimes\mathcal A_2$. Can we conclude, that if $$\left\|f(\;\cdot\;,\omega_2)\right\|_{L^1(\mu_1)}<\infty\;\;\;\text{for }\mu_2\text{-almost every }\omega_2\in\Omega_2$$ and $$\left\|f(\omega_1,\;\cdot\;)\right\|_{L^1(\mu_2)}<\infty\;\;\;\text{for }\mu_1\text{-almost every }\omega_1\in\Omega_1\;,$$ then $$\left\|f\right\|_{L^1(\mu_1\otimes\mu_2)}<\infty\;?$$

Answer 1

The answer is no. Take $\Omega_1 = \Omega_2 = \mathbb{R}$ with the usual Lebesgue measure and set $f(x_1,x_2) = e^{-(x_1-x_2)^2}$. Then $\|f(\cdot, x_2)\|_{L^1(\Omega_1)} = \|f(x_1, \cdot)\|_{L^1(\Omega_2)} = const. < \infty$ for all $x_1, x_2$ but $f \notin L^1(\mathbb{R}^2)$.