I'd like to show some properties of the Lebesgue integral.
I'd like to show that if $f$ is a simple function which is zero almost everywhere, then the Lebesgue integral $\int f(x) dx = 0$.
Similarly, I'd like to show this is also true for a measurable function $f$ which is zero almost everywhere.
I'm working through a real analysis textbook on my own, and I'm not quite sure what to do about this "almost everywhere." Do I have to consider separately a set of measure zero? Thank you as always.
Attempt for simple function:
If $f$ is a simple function that is zero almost everywhere, then $f = \sum_{i=1}^{n}a_{i}\chi_{E_i} = 0$.
Then, for each $i$, either $a_i = 0$ or else $m(E_i)= 0$.
By definition, $\int f(x) = \sum_{i=1}^{n}a_{1}m(E_i)$.
This summation is the sum of zeros. Thus, $\int f(x) = 0$ as desired.
Attempt for measurable function:
Assume $f$ is a non-negative measurable function that is zero almost everywhere.
By definition, $\int f(x) dx = \lim_{n \to \infty}\int f_n(x) dx$ where $\{f_n\}$ is a sequence of increasing, non-negative, simple functions that are all less than $f$.
Since $f$ is zero almost everywhere, then each non-negative, simple $f_n$ must also be zero almost everywhere.
Now, from above, we know that for each $n \in \mathbb{N}$, $\int f_n(x) dx = 0$.
Then, $\int f(x) dx = \lim_{n \to \infty} 0 = 0$.
Thus, $\int f(x) dx = 0$.
Now, for the general case…
We can write any measurable function $f$ in terms of its positive and negative parts. So, $f(x) = f^+(x) – f^-(x)$.
Both $f^+(x)$ and $f^-(x)$ are non-negative. Now if $f$ is zero almost everywhere, then both $f^+(x)$ and $f^-(x)$ are non-negative and zero almost everywhere.
Then, by above, $\int f^+(x) dx= \int f^-(x) dx= 0$
And by definition, $\int f(x) dx = \int f^+(x) dx – \int f^-(x) dx$.
So, $\int f(x) dx = 0 – 0 = 0$ as desired.
Best Answer
The Lebesgue integral is monotone, that is if $f$ and $g$ are integrable and $f(x)\geq g(x)$ for all $x$, then $\int f\geq\int g$.
So take any nonnegative meaurable function $f$ that is zero almost everywhere and let $S$ be the set $S=\{x:f(x)\neq 0\}$. It is easily seen that $S$ is measurable and by assumption, $S$ has measure $0$. Define a function $f'$ that takes the value $\infty$ on $S$ and $0$ everywhere else. Let $f_0$ be the function that is constantly $0$. Then $f_0(x)\leq f(x)\leq f'(x)$ for all $x$ and hence $0=\int f_0\leq\int f\leq\int f'=0$. For $\int f'=0$, we use the fact that the sequence of simple functions $(f_n)$ defined so that $f_n$ takes the value $n$ on $S$ and $0$ everywhere else is an increasing sequence converging to $f'$.
For a general measurable function, you apply the argument to both the positive part and the negative part.