# $(X,\mu)$ is a measure space and $f:X\to\mathbb R$ be non-negative integrable function vanishing at a measure $0$ set then is $\int f>0$?

Let $(X,\mu , \mathcal F)$ be a measure space , $f:X\to\mathbb R$ be a non-negative measurable function such that $\mu (f^{-1}\{0\})=0$ and $f$ is integrable , then is it true that $\int_X f d\mu >0$ ?

Let $A_n=\left \{ x\in X:f(x)>\frac{1}{n} \right \}$ and assume $\mu (X)\neq 0$.
Then $A_n\subseteq A_{n+1},\ \bigcup_{n}A_n=X\setminus f^{-1}\{0\}$ and $\lim_{n\to \infty } \mu (A_n)=\mu (X)$.
Thus, there is an $N\in \mathbb N$ such that $\mu (A_N)>0$.
But then, $\int_X f d\mu\ge \int_{A_N} f d\mu>0.$