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Let $f,g:S\rightarrow\mathbb{R}$; assume $f$ and $g$ are integrable over $S$. Show that if $f$ and $g$ agree except on a set of measure zero, then $\int_Sf=\int_Sg$.

Since $f$ and $g$ are integrable over $S$, we have $f-g$ also integrable over $S$. So $(f-g)(x)=0$ except for a set of measure zero. If $(f-g)$ were bounded, we could choose a partition so that the volume covering the points $x$ such that $(f-g)(x)\neq 0$ is less than any $\epsilon$, which will imply that $\int_S(f-g)=0$, and so $\int_S f=\int_S g$. But here we don't have boundedness. How can we go from here?

Mika H.
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    The integral of any measurable function over any set of measure zero is always equal to zero (if you take the convention $\infty\cdot 0=0=0\cdot\infty$ which is standard in measure theory). – Amitesh Datta Aug 31 '13 at 04:49
  • Judging from your comment, I would guess that you are not taking standard measure theory course. What is your definition of a measure zero set? – Vishal Gupta Aug 31 '13 at 05:11
  • @Vishal A measure zero subset of $\mathbb{R}^n$, given any $\epsilon$, can be covered by a countable collection of rectangles such that the sum of volumes is less than $\epsilon$. – Mika H. Aug 31 '13 at 05:16

3 Answers3

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If you want to go with all the details starting from the definition, do it like this : $$ \left| \int_S (f - g) \, d\mu \right| \le \int_S |f-g| d\mu = \sup \left\{ \left. \int_S \varphi \, d\mu \, \right| \, 0 \le \varphi \le |f-g|, \varphi \in \mathcal L(S)\right\} $$ where I wrote $\mathcal L(S)$ for the set of all linear combinations of characteristic functions (a characteristic function is a function which is $1$ on some measurable set and $0$ elsewhere). Since for these functions, the integral is defined as $$ \int_S \left( \sum_{i=1}^n a_i \mathbb 1_{A_i} \right) d\mu = \sum_{i=1}^n a_i \mu(A_i), $$ the condition $0 \le \varphi \le |f-g|$ ensures that $a_i \neq 0 \implies \mu(A_i) = 0$, hence $\int_S \varphi d\mu = 0$ for every $\varphi$ with $0 \le \varphi \le |f-g|$. Taking the supremum over a bunch of zeros is zero.

Hope that helps,

Patrick Da Silva
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The integral of any measurable function over any set of measure zero is equal to zero. You needn't worry about boundedness in your question because the definition of "Riemann integrability" entails boundedness and thus your argument is complete. Let me know if you need further clarification on this point and I'm happy to discuss it with you further!

Amitesh Datta
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    Amitesh, the book I'm studying hasn't mentioned "measurable function" yet. The book defines integrals on a rectangle (based on partitions), integrals on a bounded set (based on using a rectangle to cover it), and sets of measure zero (used to specify condition that integral exists). Perhaps you could clarify measurable function a little bit? (Or refer to something more elementary that is enough for this problem.) – Mika H. Aug 31 '13 at 05:05
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    Sure, @Mika. The key point is that Riemann integrable functions are assumed to be bounded (boundedness is a part of the definition of "Riemann integrable"). A set of measure zero can be covered by rectangles of total volume ($=$ the sum of the volumes of the rectangles) arbitrarily small, by definition. Thus, the Riemann integral of a Riemann integrable function over a set of measure zero must equal to zero. Do you think that answers your question? – Amitesh Datta Aug 31 '13 at 05:17
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    Amitesh, yes I think so. I should have realized, but didn't, that boundedness is implicit in the Riemann integral definition. So I don't need to worry about it here. – Mika H. Aug 31 '13 at 05:29
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    Thanks, @Mika for accepting my answer. BTW, when you do get to studying Lebesgue integration and general measure theory, it's worth keeping in mind that general measurable functions needn't be bounded! (The theory of Lebesgue integration is far more flexible than the theory of Riemann integration.) – Amitesh Datta Aug 31 '13 at 05:35
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    Thanks Amitesh. I will keep that in mind! – Mika H. Aug 31 '13 at 05:42
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You do not need boundedness to conclude that integral of a function over a set of measure zero is zero. It is true for all measurable functions, bounded or not.

To see this, let $f$ be a function which is zero almost everywhere (i.e. except on a set of measure zero). Then let

$$ E = \{ x \in X \; | \; f(x) > 0 \}$$ $$ F = \{ x \in X \; | \; f(x) < 0 \}$$

Clearly both $E$ and $F$ are of sets of measure zero.

Now let $\phi_{n}$ be a sequence of simple functions such which increase to $f\chi_{E}$ pointwise. Then $\forall n \in \mathbb{N}, \phi_{n}$ is zero except possibly on $E$ which is of measure zero. And hence $\int_{X}\phi_{n} = 0\; \forall n$ and hence $\int f\chi_{E} = 0$ and similarly, $\int f\chi_{F} = 0$ and hence $\int_{X}f = 0$.

Vishal Gupta
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