## 7.4. Lebesgue Integral

### Example 7.4.6(a): Lebesgue Integral for Bounded Functions

We have shown before that the function*f(x) = x*is Riemann integrable. Since step functions are special cases of simple functions, we could try to replicate that proof here, suitably modified for the function

^{2}*f(x) = x*. But using simple functions instead of step functions actually simplifies the proof, so here we go.

We know that *| f(x) | 1* over
the interval *[0, 1]*. Define sets:

forE_{1}= { x [0, 1]: 0 f(x) < 1/n }

E_{2}= { x [0, 1]: 1/n f(x) < 2/n }

E_{3}= { x [0, 1]: 2/n f(x) < 3/n }

...

E_{j}= { x [0, 1]: (j-1)/n f(x) < j/n}

*j = 1, 2, ..., n*. Because

*f*is continuous, the sets

*E*are measurable (really - why?), they are disjoint, and their union (over the

_{j}*j*'s) equals

*[0, 1]*(actually, the union equals

*[0, 1)*, but that does not matter - why?).

Now define two simple functions

Fix an integerS_{n}(x) = j/n X_{Ej}(x)

s_{n}(x) = (j-1)/n X_{Ej}(x)

*n*and take a number

*x*in

*[0, 1)*. Then

*x*must be contained in exactly one set

*, and on that set we have*

**E**_{j}Therefore, on all ofs_{n}(x) = (j-1)/n f(x) < j/n = S_{n}(x)

*[0, 1]*, we know that

But thens_{n}(x) f(x) S_{n}(x)

ThereforeI^{*}(f)_{L}S_{n}(x) dx = 1/n j m(E_{j})

I_{*}(f)_{L}s_{n}(x) dx = 1/n (j-1) m(E_{j})

SinceI^{*}(f)_{L}- I_{*}(f)_{L}1/n (j - (j-1)) m(E_{j}) =

= 1/n m(E_{j}) = 1/n m([0, 1]) = 1/n

*n*was arbitrary the upper and lower Lebesgue integrals must agree, hence the function

*f*is integrable.

Note that - with a few simple modifications - this proof could show that
*every bounded function* *f* which has the property that
*the sets E_{j} are measurable*
is Lebesgue integrable.

It remains, though, to find the actual value of the integral. But we can
easily compute the measure of the sets *E _{j}* using the
fact that

*f(x) = x*: for a fixed

*n*we have

From the above computation it follows thatm( E_{j}) = m({ x [0, 1]: (j-1)/n f(x) < j/n}) =

= m({ x [0, 1]: (j-1)/n x < j/n}) =

= m( [(j-1)/n, j/n] ) = 1/n

which is, incidentally, the same value as for the Riemann integral.f(x) dx = lim 1/n j m(E_{j}) =

= lim 1/n j 1/n = = lim 1/n^{2}j =

= lim 1/n^{2}1/2 n (n-1) = 1/2