College Math Teaching

January 18, 2011

Lebesgue Integral: Bounded Functions on a Bounded Set

Now that we have an idea of what the Lebesgue integral is, how do we define it?

If we limit ourselves to bounded, measurable functions on the real line, we could do the following:
suppose there is a real number M>0 such that -M\leq f\leq M over [0,1]. Then for some integer k, we could set up the partition of the range:

-M=y_{-k},\frac{(1-k)M}{k}=y_{-k+1},\frac{(2-k)M}{k}=y_{-k+2},....0=y_{0},\frac{M}{k}=y_{1},....\frac{(k-1)M}{k}=y_{k-1},M=y_{k}

Now set up E_{-k}=f^{-1}([y_{-k},y_{-k+1})),....E_{k}=f^{-1}([y_{k-1},y_{k}))

And then have \varphi _{k}=\sum_{i=-k}^{k-1}y_{i}m(E_{i}) and \psi_{k}=\sum_{i=-k+1}^{k}y_{i}m(E_{i})

Note that \varphi _{k} plays the role of the lower sum and \psi _{k} plays the role of the upper sum. If \inf \psi _{k}=\sup \varphi _{k} the function is integrable (as it always is if f is measurable and bounded) and we define \int_{0}^{1}f(x)dx=\inf \psi _{k}=\sup \varphi _{k}.

Note: It is possible to define the Lebesgue integral without having the concept of measurable function first: we can start with a bounded function
f and partition the range of f by y_{0},y_{1},...y_{n}. We can then look at all partitions E_{1}....E_{n} of [0,1] by measurable sets.

Then consider the characteristic functions \chi _{i}(x)=\left\{ \begin{array}{c}1,x\in E_{i} \\ 0,x\notin E_{i}\end{array}\right. so we can form a type of general step function \varphi_{n}=\sum_{i=1}^{n}y_{i-1}m(E_{i}) and \psi_{n}=\sum_{i=1}^{n}y_{i}m(E_{i}) and call f integrable if \sup \{\varphi_{n},\varphi _{n}\leq f\}=\inf \{\psi _{n},\psi _{n}\geq f\}. Think of the first function as approximating f from below by generalized step functions, and the second as approximating f from above.

Note: one of the big deals about the Lebesgue integral is that we get better convergence properties; that is, if we have a sequence of integrable
functions $f_{n}\rightarrow f$ pointwise over a measurable set, then with only mild extra hypothesis, we can show that the limit function is also
integrable and that the integral can be obtained as some sort of limit of integrals.

But to make any headway on such theorems, we’ll have to retreat to some theorems concerning measurable sets; so far we’ve shown that some sets are not measurable; we haven’t developed sufficient conditions for a set to be measurable.

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1 Comment »

  1. […] Lebesgue Integration and measure I Lebesgue Integration and measure 2 Lebesgue Integration and measure 3 Lebesgue Integration and measure 4 […]

    Pingback by 18 January 2011: posting « blueollie — January 18, 2011 @ 12:45 pm


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