## Example 7.4.7(d): Riemann implies Lebesgue Integrable

Show that the restriction of a bounded continuous function to a measurable
set is Lebesgue integrable.

We should rephrase the question slightly so that it makes sense: suppose
*f*is a bounded continuous function over an interval

*(a, b)*and

*is a measurable subset of*

**A***(a, b)*with finite measure. Then

*f*is Lebesgue integrable over

*.*

**A**
First note that if *[c, d)* is an interval contained in
*(a, b)* then the set *f ^{ -1 }([c, d))* is
measurable, because:

and[c, d) = (c-1/n, d)

Sincef^{ -1 }([c, d)) = f^{ -1 }( (c-1/n, d)) = f^{ -1 }((c-1/n, d))

*f*is continuous, each set

*f*is open, hence measurable, and the intersection of measurable sets is again measurable.

^{ -1 }((c-1/n, d))
Now suppose *f* is bounded by *M* and fix an integer
*n*. Define the sets

whereE_{j}= { x (a, b): (j-1)M/n f(x) < jM/n }=A

= f^{ -1 }( [(j-1)M/n, jM/n) )A

*-n j n*. Because of above each set

*E*is measurable. Now the proof can continue just as in a previous exercise, so we will only sketch it here.

_{j}
**First:** Define functions

where the sum is taken froms_{n}(x) = M/n (j-1) X_{Ej}(x)

S_{n}(x) = M/n j X_{Ej}(x)

*j = -n*to

*n*. Then the sets

*E*are measurable, disjoint, and their union is

_{j}*, and the functions*

**A***s*and

_{n}*S*are simple and integrable.

_{n}
**Second:** Verify that

(1)s_{n}(x) f(x) S_{n}(x)(2)

Iand^{*}(f)_{L}S_{n}(x) dxI_{*}(f)_{L}s_{n}(x) dx(3)

I^{*}(f)_{L}- I_{*}(f)_{L}M/n m()A

**Third:**Finish the proof.