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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.2012-05-13
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    Welcome to MSE! What book are using?2012-05-13
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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?2012-05-13
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    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!2012-05-13
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    You have to assume that the simple function in the definition of the integral converge pointwise to $f$.2012-05-13

1 Answers 1

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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?2012-05-13
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    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.2012-05-13
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    I'm sorry. I don't quite follow. Could you explain a little further? Thanks for your help!2012-05-13
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    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 $0$.2012-05-13
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    Are all the simple functions zero (a.e.) because they are non-negative and less than $f$ which is zero (a.e.)?2012-05-14
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    Yes. The set of points on which they are not equal to zero must be a subset of the points at which $f$ is equal to zero.2012-05-14
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    I've edited my attempt. Thoughts?2012-05-14
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    It's correct now.2012-05-14
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    You mentioned that we have to consider both the negative and the positive parts for the general case. I've included so in an edited proof. What do you think? However, I don't really understand why we have to do this. If the function is zero almost everywhere then why do we even consider negative parts?2012-05-14
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    The Lebesgue integral is first *defined* for nonnegative meaurable functions and then extended by taking positive and negative parts, so that i simply the definition. And clearly, a function zero almost everywhere can take negative values on a set of measure zero.2012-05-14
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    So, if we take any measurable function that is zero (a.e.) and split it into it's positive and negative parts, will both $f^+(x)$ and $f^-(x)$ both be non-negative and zero (a.e.) as I have indicated in my proof? Thank you so much for your help. I'm self-teaching myself and without help from kind individuals like yourself I'm not sure I'd ever understand.2012-05-14
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    Yes, that is correct. The reason is that $\{x:f(x)>0\}\subseteq \{x:f(x)\neq 0\}$ and $\{x:f(x)<0\}\subseteq\{x:f(x)\neq 0\}$ and a measurable subset of a set with measure zero has measure zero too.2012-05-14