Real-Analysis-The General Lebesgue Integral, guaranteed and verified 100% Pass

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The General Lebesgue Integral


If 𝑓 is an extended real valued function on 𝐸 define:

𝑓 + (𝑥 ) = max⁡{𝑓(𝑥 ), 0} ≥ 0
𝑓 − (𝑥 ) = max⁡{−𝑓(𝑥 ), 0} ≥ 0
𝑓(𝑥 ) = 𝑓 + (𝑥 ) − 𝑓 − (𝑥 )
Notice that 𝑓 is measurable if and only if 𝑓 + and 𝑓 − are measurable.



Prop. Let 𝑓 be a measurable function on 𝐸. Then 𝑓 + and 𝑓 − are integrable
over 𝐸 if and only if |𝑓| is integrable over 𝐸.



Proof: Assume 𝑓 + and 𝑓 − are integrable over 𝐸.

Notice that |𝑓 | = 𝑓 + + 𝑓 − .

Thus ∫𝐸 |𝑓 | = ∫𝐸 𝑓 + + ∫𝐸 𝑓 − < ∞,

and |𝑓| is integrable over 𝐸.


Now assume |𝑓| is integrable over 𝐸.

0 ≤ 𝑓 + ≤ |𝑓| and 0 ≤ 𝑓 − ≤ |𝑓| .

So by monotonicity: ∫𝐸 𝑓 + ≤ ∫𝐸 |𝑓| < ∞, ∫𝐸 𝑓 − ≤ ∫𝐸 |𝑓| < ∞.
Thus 𝑓 + and 𝑓 − are integrable over 𝐸.

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Def. A measurable function 𝑓 on 𝐸 is said to be integrable over 𝑬 if |𝑓| is
integrable over 𝐸. When this is so we define:

∫𝐸 𝑓 = ∫𝐸 𝑓 + − ∫𝐸 𝑓 − .

Notice if 𝑓 is nonnegative, i.e. 𝑓 = |𝑓 | = 𝑓 + , ⁡𝑓 − = 0 and we get the usual
definition of the Lebesgue integral of a nonnegative function. If 𝑓 is a bounded
measurable function of finite support by linearity of integration this definition
coincides with the original definition.

Notice also that, unlike Riemann integration, in order for a function 𝑓 to be
Lebesgue integrable we require |𝑓| to also be integrable.


𝑠𝑖𝑛𝑥
Ex. 𝑓 (𝑥 ) = is integrable as a Riemann integral over [0, ∞), but not as a
𝑥
𝑠𝑖𝑛𝑥
Lebesgue integral because ∫[0,∞) | | = ∞.
𝑥



Prop. Let 𝑓 be integrable over 𝐸. Then 𝑓 is finite a.e. on 𝐸 and

∫𝐸 𝑓 = ∫𝐸~𝐴 𝑓 if 𝐴 ⊆ 𝐸 and 𝑚(𝐴) = 0.


Proof: We know if 𝑔 is nonnegative and 𝑔 is integrable over 𝐸 then 𝑔 is finite
a.e. on 𝐸. Thus |𝑓| is finite a.e. on 𝐸 and hence 𝑓 is.


Since 𝑓 is integrable: ∫𝐸 𝑓 = ∫𝐸 𝑓 + − ∫𝐸 𝑓 −

and ⁡⁡∫𝐸 𝑓 + = ∫𝐸~𝐸 𝑓 + ∫𝐸 𝑓 − = ∫𝐸~𝐸 𝑓 − .
0 0



So ∫𝐸~𝐸 𝑓 = ∫𝐸~𝐸 𝑓 + − ∫𝐸~𝐸 𝑓 − = ∫𝐸 𝑓 + − ∫𝐸 𝑓 − = ∫𝐸 𝑓.
0 0 0