Let $ f $ be a nonnegative measurable function defined on a measurable set $ E $. Define its Lebesgue intergal as the supremum of the set $ \{\int_{E} h \} $ for all bounded, measurable functions $ h $ such that $ 0 \leq h \leq f $ and that $ m\{x \in E | h(x) \neq 0 \} $ is finite.
Theorem: (Proposition 9 Chapter 4 in Real Analysis 4th edition by Royden). The function $ f = 0 $ almost everywhere on $ E $ iff $ \int_{E} f = 0 $.
Proof: (Ideas extracted from Royden). First suppose $ \int_{E} f = 0 $. For every $ n \in \mathbb{N}^{+}$, by Chebyshev Theorem $ 0 \leq m\{x \in E | f(x) \geq 1 /n \} \leq \frac{1}{n} \int_{E} f = 0 $.
So $ m\{x \in E | f(x) \geq 1 /n \} = 0 $ for all $ n $, so $ f = 0 $ almost everywhere.
Now suppose $ f = 0 $ almost everywhere.
Let $ \phi $ be a simple function and $ h $ be a bounded measurable function such that $ m\{x \in E | h(x) \neq 0 \} $ is finite and $ 0 \leq \phi \leq h \leq f $.
Then $ \phi = 0 $ almost everywhere since $ m\{x \in E | \phi(x) \neq 0 \} \leq m\{x \in E | f(x) \neq 0 \} = 0 $.
Hence $ \int_{E} h = 0 $.
Consequently $ \int_{E} f = 0 $.
My question is why does $ \phi = 0 $ almost everywhere implies $ \int_{E} h = 0 $.
Am I missing anything here?
If yes, how should I approach the converse direction
Best Answer
There is a conceptually easier way to prove the converse if we use the usual (at least, what Wikipedia says) definition of the Lebesgue integral, which is just $\int_E f = \sup\{\int_E h \mid h \ \text{simple}, \ h \leq f\}$. For then if $f = 0$ a.e., every nonnegative simple function $h$ approximating $f$ from below is zero a.e. and is thus of the form $h = \sum_{i = 1}^n a_i\chi_{A_i}$, where either $a_i = 0$ or $m(A_i) = 0$. Then $\int_E h = \sum_{i = 1}^n a_im(A_i) = 0$. So $\int_E f = \sup\{\int_E h \mid h \ \text{simple}, \ h \leq f\} = 0$.