Page 378 - DMTH401_REAL ANALYSIS
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Real Analysis
Notes + -
we cannot always do that, because it may well happen that f ò and f ò are both infinite, in
which case their difference would be meaningless. (Remember that – is undefined.) So we
need to restrict ourselves to a smaller class of functions than the collection of all measurable
functions when we drop the non-negative requirement and come to the following definition.
Definition: For f : E [–, ], denote f+ = f V0 and f = (–f) V0. Then f is said to be integrable if
–
+
-
E ò
and only if both ò f and f are finite, in which case we define the integral of f by
E
A ò f = A ò f + A ò - f -
for any A E
Notation: We shall denote the class of all (extended real-valued) integrable functions defined on
E by C(E).
+
Note that in the above definition, f and f are both non-negative measurable, so for any set
–
+ - + + -
A E, ò f and ò f are both defined. Furthermore, ò f £ E ò f < and similarly ò f < so
A A A A
their difference makes sense now. Also note that for non-negative integrable functions this
definition agrees with our old one.
We provide an alternative characterization of integrable functions.
Proposition: A measurable function f defined on E is integrable if and only if |f|< so.
E ò
–
Proof: Just note that |f| = f + f .
+
We proceed to investigate the structure of (E). We want to say it is a vector lattice. But we have
to be careful here: Given f, g Î (E) it may well happen that f(x) = + and g(x) = – for some x
Î E and then f + g cannot be defined by f(x) + g(x) at that x. Luckily there cannot be too many such
x’s, in the sense that the set of all such x’s is of measure zero. In fact every integrable function is
finite. We know that the values of a function on a set of measure zero are not important as far as
integration is concerned. (This was observed as in the case of bounded measurable functions
vanishing outside a set of finite measure; the reader should verify this for the case of general
integrable functions as well.) So that eliminates our previous worries: more precisely, let us
agree from now on two functions f,g: E [–, ] are said to be equal (write f = g) if and only if
they take the same values a.e.on E, and f + g shall mean a function whose value at x is equal to f(x)
+ g(x) for a.e.x Î E. Also say f £ g if and only if f(x) £ g(x) for a.e. x Î E. Then we have the following
proposition.
Proposition: (E) forms a vector lattice (partially ordered by £).
£
+
E ò
Proof: If f,g Î (E), then |f g| ò E |f|+ E ò |g|< (we are using linearity and monotonicity and
hence f + g Î (E) (the measurability of f + g is previously known). The rest of the proposition
is trivial.
With the vector lattice structure of (E) it is natural to ask whether the integral is linear and
monotone or not. We expect it to be true; we verify it below.
+
f
Proposition: For any f,g Î (E) and A E, we have ò A (f g) = A ò f + A ò g and ò = A ò f .
A
Furthermore, if f £ g a.e. on A then ò A f £ A ò g .
f
Proof: The parts for monotonicity and ò = f are easy and left as an exercise.
A A ò
+
So now let f,g Î (E) and A E be given, and we prove ò (f g) = A ò f + A ò g . By definition of the
A
-
-
+
-
integral, the LHS is just (f g)+ + - A ò (f g) and the RHS is ò A f - A ò f + A ò g + A ò g , all terms being
+
A ò
finite. So it suffices to show
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