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I am having some difficulty in understanding notation/meaning in the definition of definite integral. Can you guys help clarify/correct and fill in the gaps in my understanding.

Following is from wikipedia entry for the Darboux integral which is similar to what is in my textbook. My notes are inline.

A partition of an interval $[a,b]$ is a finite sequence of values $x_i$ such that,

$a = x_0 < x_1 < ... < x_n = b $

Partition to integrate over

Each interval $[x_i−1,x_i]$ is called a subinterval of the partition. Let $ƒ:[a,b]→R$ be a bounded function, and let

$P = (x_0, ..., x_n)$

be a partition of $[a,b]$. Let

Defining rectangles that would be used to approximate area under curve

$M_i = \sup\limits_{x \in [x_{i-1}, x_i]} f(x)$

$m_i = \inf\limits_{x \in [x_{i-1}, x_i]} f(x)$

Height of rectangle when above the curve or below the curve respectively

The upper Darboux sum of ƒ with respect to P is

$U_{f,P} = \sum_{i=1}^n (x_i - x_{i-1})M_i$

sum when using supremum for height

The lower Darboux sum of ƒ with respect to P is

$L_{f,P} = \sum_{i=1}^n (x_i - x_{i-1})m_i$

sum when using infimum for height

The upper Darboux integral of ƒ is

$U_f = \inf \{U_{f,P} : \mbox{P is a partition of} [a,b]\}$

The lower Darboux integral of ƒ is

$L_f = \sup \{L_{f,P} : \mbox{P is a partition of} [a,b]\}$

This is the part that I am unable to understand. What do the infimum and supremum imply here? I understand the definite integral as area under curve. Wouldn't the sums $U_{f,P}$ and $L_{f,P}$ then be the integrals themselves? Why do we need to take the infimum/supremum of these areas?

If $U_ƒ = L_ƒ$, then we say that $ƒ$ is Darboux-integrable and set

$\int_a^b f(t)dt = U_f = L_f$

this parts seems to intuitively suggest both areas would be nearly equal

Thanks for your help.

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1  
The upper and lower Darboux sums only approximate the area under the curve $(x, f(x))$. There is no single partition $P$ of $[a, b]$ such that $U_{f,P}$ is the exact area, but the approximation keeps getting better as $P$ gets finer. An equivalent way to say that $f$ is integrable is for every $\epsilon > 0$, there is a partition $P$ of $[a, b]$ such that $U_{f, P} - L_{f, P} < \epsilon$. –  Vectk Aug 10 '12 at 7:32
    
Thank you, I understand now. –  mathguy80 Aug 10 '12 at 7:39

2 Answers 2

up vote 1 down vote accepted

We take infimum and supermum with respect to partitions. Note that $U_{f,P}$ and $L_{f,P}$ depend on $P$. For each $P$, we know that $L_{f,P} \le U_{f,P}$. As $P$ gets finer, $U_{f,P}$ may decrease and $L_{f,P}$ may increase. If there exists a sequence of partition $P_1, P_2, \ldots$ that makes $U_{f,P_n}$ and $L_{f,P_n}$ converge to the same number, that number is taken as the integral. We require that the sequence of partitions must be "increasing" in the sense that all grid points in $P_n$ must be in $P_{n+1}$.

However, there are infinitely many choices of increasing sequences of partitions. If we find a sequence that "doesn't work", that doesn't mean there doesn't exist a sequence that works. To take into account ALL partitions at once, $\sup$ and $\inf$ are used. Note that not only we know that $L_{f,P} \le U_{f,P}$, but we actually know that $L_{f,P} \le U_{f,Q}$ for any partitions $P$ and $Q$. That means every number in the set $\{L_{f,P}|P\text{ is a partition}\}$ is smaller than or equal to every number in the set $\{U_{f,P}|P\text{ is a partition}\}$. If the two set has a common accumulation point, then we can create a sequence of partitions $P_1, P_2, \ldots$ such that $U_{f,P_n}$ and $L_{f,P_n}$ converge to the same number. (Pick once sequence for $U_f$ and one sequence for $L_f$, then create a new partition sequence by taking the union of partitions from both sequences.)

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Thanks. I ignored the idea that there multiple partitions. –  mathguy80 Aug 10 '12 at 7:38

$U_{f,P}$ is the area of the upper rectangles and is, generally, bigger than the area under $f$; $L_{f,P}$ is the area of the lower rectangles and is, generally, less than the area under $f$. $U_f$ is the infimum of $U_{f,P}$ over all possible partitions, and $L_f$ is the supremum of $L_{f,P}$ over all possible partitions. If they are equal, then they must equal the integral.

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Thanks. Makes sense now. –  mathguy80 Aug 10 '12 at 7:37

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