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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.

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Welcome to Math.stackexchange! Here are some starters: A simple function has a range which is a finite set - they can be expressed in a 'simple' way. Can you tell us what that is? From there, you want to split it up into two parts - the non-zero part and the zero part. Then, tell us what your definition for the Lebesgue integral for a simple function is, and apply that definition. –  Ragib Zaman May 13 '12 at 1:52
    
Welcome to MSE! What book are using? –  leo May 13 '12 at 3:35
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Some more hints: 1. Reduce to problem to characteristic functions. 2. What does it mean that $f$ is a characteristic funtion wich is $0$ almost everywhere? 3. How is defined the integral of a characteristic function? –  leo May 13 '12 at 3:41
    
I've added an attempt to the question. Any hints, advice, our direction is much appreciated. Also, I'm using Krantz's Real Analysis and Foundations text. Thank you for your warm welcome. This looks to be a great community! –  J. Chong May 13 '12 at 20:56
    
You have to assume that the simple function in the definition of the integral converge pointwise to $f$. –  Michael Greinecker May 13 '12 at 23:33

1 Answer 1

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.

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I'd like to show the result without using the fact that the Lebesgue integral is monotone--as this is a result I haven't yet proven. Any advice for that sort of a solution? –  J. Chong May 13 '12 at 23:19
    
For simple functions, your attempt above is correct. For general functions. For nonnegative measurable functions, show that if $f$ is zero almost everywhere and $f_s$ is a simple function not larger at any point, then $f_s$ is zero almost everywhere. Apply the definition of the Lebesgue integral. For the general case, take positive and negative parts. –  Michael Greinecker May 13 '12 at 23:30
    
I'm sorry. I don't quite follow. Could you explain a little further? Thanks for your help! –  J. Chong May 13 '12 at 23:49
    
The Lebesgue integral of a nonnegative mesurable function $f$ is the limit of integrals of a sequence that converges to the function from below. If you can show that all this simple functions are zero almost everywhere, you know they have all integral $0$ and therefore the integral of $f$ is $0$. –  Michael Greinecker May 13 '12 at 23:52
    
Are all the simple functions zero (a.e.) because they are non-negative and less than $f$ which is zero (a.e.)? –  J. Chong May 14 '12 at 0:20

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