Let $(E,\mathcal E,\mu)$ be a measure space and $f:E^2\to[0,\infty)$ be $\mathcal E^{\otimes2}$-measurable with $$f(x,y)>0\;\;\;\text{for }\mu^{\otimes2}\text{-almost all }(x,y)\in E^2.\tag1$$
Are we able to conclude that $$g(x,y):=\min(f(x,y),f(y,x))>0\;\;\;\text{for }\mu^{\otimes2}\text{-almost all }(x,y)\in E^2\text?\tag2$$
It would be trivial, if we would have $$f(x,y)>0\;\;\;\text{for }\mu\text{-almost all }x,y\in E\tag{1'}$$ instead of $(1)$, but with $(1)$ I'm unsure how we need to argue. The problem is that by assumption there is a $\mu^{\otimes2}$-null set $N$ with $f(x,y)>0$ for all $(x,y)\in E^2\setminus N$, but $(x,y)\in E^2\setminus N$ doesn't imply $(y,x)\in E^2\setminus N$.
If $N$ is a $\mu^{\otimes2}$-null set then so are $N'=\{(y,x)\mid (x,y)\in N\}$ (by symmetry) and $M=N\cup N'$.
And if: $$f(x,y)\leq 0\implies (x,y)\in N$$ then: $$f(y,x)\leq 0\implies (x,y)\in N'$$ and: $$g(x,y)=\min(f(x,y),f(y,x))\leq0\implies (x,y)\in M$$