Lebesgue integral question:

Consider measure space $\displaystyle (X, \epsilon, \lambda)$ with $\displaystyle f$ a summable function.

I want to show that if $\displaystyle \int_{X}|f|d\lambda = 0$ then $\displaystyle |f| = 0$ a.e. and therefore $\displaystyle f = 0$ a.e.

My proposed proof is as follows:

Assume that $\displaystyle |f| > 0$ for some $\displaystyle x \in X$.

Since $\displaystyle |f|$ is positive measurable function it follows that it is the limit of an increasing sequence of simple functions which I will show as $\displaystyle |f| = \sum_{i=1}^{k}C_{i,n}\chi_{E_{i},n}$:

Using Beppoi Levi I know that the limit of the integral is the integral of the limit which gives the following:

$\displaystyle \int_{X}|f|d\lambda = \int_{X}\lim\limits_{n \rightarrow \infty}\sum_{i=1}^{k}C_{i,n}\chi_{E_{i},n} = \lim\limits_{n \rightarrow \infty}\int_{X}\sum_{i=1}^{k}C_{i,n}\chi_{E_{i},n} \\ = \lim\limits_{n \rightarrow \infty}\sum_{i=0}^{k}c_{i,n}\lambda(E_{i,n}) = 0$

This is only possible if for any $\displaystyle n \in \mathbb{N}$ we have $\displaystyle \sum_{i=1}^{k}c_{i,n}\lambda(E_{i,n}) = 0$.

$\displaystyle \therefore$ for all $\displaystyle n$ and $\displaystyle i$ it follows that either $\displaystyle c_{i,n} = 0$ or $\displaystyle \lambda(E_{i,n}) = 0$.

Since we assumed that $\displaystyle |f|$ is nonzero for some $\displaystyle x \in X$ it follows that there is some $\displaystyle c_{i,n}\chi_{E_{i,n}} > 0$ where $\displaystyle c_{i,n} > 0$ therefore there must be some $\displaystyle E_{i,n}$ such that $\displaystyle \lambda(E_{i,n}) = 0$ and which gives $\displaystyle f(x) > 0$ for $\displaystyle x \in E_{i,n}$.

It follows that $\displaystyle |f| = 0$ a.e. and therefore $\displaystyle f = 0$ a.e.

Is this proof fine? Thanks for assistance.