Question 11.T.2: (i) Let E and F be two measurable and disjoint subsets of Ω....
(i) Let E and F be two measurable and disjoint subsets of Ω. If f ∈ \mathcal{L^{1}}(E) ∩ \mathcal{L^{1}}(F ), then f ∈ \mathcal{L^{1}}(E ∪ F ), and
\int_{E∪F} {f}dm = \int_{E}{f}dm + \int_{F}{f}dm (11.13)
(ii) If f ∈ \mathcal{L^{1}}(Ω) and g = f (a.e) on Ω, then g ∈ \mathcal{L^{1}}(Ω) and
\int_{Ω} {g}dm = \int_{Ω} {f}dm.
(iii) If f ∈ \mathcal{L^{1}}(Ω) then
m{x ∈ Ω : |f(x)| = ∞} = 0.
(iv) If f, g ∈ \mathcal{L^{1}}(Ω), then f + g ∈ \mathcal{L^{1}}(Ω) and
\int_{Ω}{(f + g)}dm = \int_{Ω}{f }dm + \int_{Ω}{g}dm. (11.14)
(v) If f ∈ \mathcal{L^{1}}(Ω) and c ∈ \mathbb{R}, then cf ∈ \mathcal{L^{1}}(Ω) and
\int_{Ω}{cf}dm = c\int_{Ω}{f }dm. (11.15)
(vi) If f, g ∈ \mathcal{L^{1}}(Ω) and f(x) ≥ g(x) (a.e.) on Ω, then
\int_{Ω}{f }dm ≥ \int_{Ω}{g }dm.
(vii) If f(x) ≥ 0 (a.e.) on Ω and \int_{Ω}{f }dm = 0, then
f(x) = 0 (a.e.) on Ω.
(viii) If f ∈ \mathcal{L^{1}}(Ω) and
\int_{E}{f }dm = 0 for all E ∈ \mathcal{M}, E ⊆ Ω,
then f(x) = 0 (a.e.) on Ω.
Note that property (ii) allows us to assume that any integrable function on Ω does not assume the values ±∞, while (iv), (v) and (vi) express the linearity and monotonicity of the Lebesgue integral.
(i) Since E ∩ F = ∅,
χ_{E∪F} = χ_{E} + χ_{F}
(f χ_{E∪F} )^{+} = f^{+}χ_{E} + f^{+}χ_{F}
\int_{Ω}{(f χ_{E∪F} )^{+}} dm = \int_{Ω}{f^{+}χ_{E}} dm + \int_{Ω}{f^{+}χ_{F}} dm
= \int_{E}{f^{+}} dm + \int_{F}{f^{+}} dm (11.16)
Similarly,
\int_{Ω}{(f χ_{E∪F} )^{−}} dm = \int_{Ω}{f^{−}χ_{E}} dm + \int_{Ω}{f^{−}χ_{F}} dm
= \int_{E}{f^{−}} dm + \int_{F}{f^{−}} dm (11.17)
Consequently, f χ_{E∪F} is integrable on Ω, that is, f is integrable on E ∪ F, if and only if f is integrable on E and F, and (11.13) follows by subtracting (11.17) from (11.16).
(ii) Note first that g ∈ \mathcal{L^{0}}(Ω) by Theorem 10.19. Setting A = {x ∈ Ω : f(x) ≠ g(x)}, we see that m(A) = 0 and Ω \setminus A ∈ \mathcal{M}. Using Remarks 11.2.2 and 11.2.3, we conclude that f ∈\mathcal{ L^{1}}(Ω \setminus A) and
\int_{A}{f} dm = \int_{A}{g} dm = 0.
Since gχ_{Ω \setminus A} = f χ_{Ω \setminus A}, g ∈\mathcal{ L^{1}}(Ω \setminus A) and
\int_{Ω \setminus A}{g} dm = \int_{Ω \setminus A}{f} dm.
Using (i) we therefore have g ∈ \mathcal{L^{1}}(Ω) and
\int_{Ω}{g} dm = \int_{Ω \setminus A}{g} dm + \int_{A}{g} dm
= \int_{Ω \setminus A}{f} dm + \int_{A}{f} dm
= \int_{Ω}{f} dm.
(iii) Setting A = {x ∈ Ω : |f(x)| = ∞} and \varphi_{n} = nχ_{A}, we see that \varphi_{n} ∈ \mathcal{L_{+}}(Ω) and \varphi_{n} \nearrow |f | χ_{A}. By Theorem 11.1,
\int_{A} |f | dm = \underset{n→∞}{\lim} n · m(A).
Since f ∈ \mathcal{L^{1}}(A), \int_{A} |f | dm < ∞ and therefore m(A) = 0.
(iv) Using part (iii), we may assume that both f and g are finite-valued functions, so that their sum is well defined on Ω. Since
(f + g)^{+} ≤ |f | + |g| , (f + g)^{−} ≤ |f | + |g| ,
we conclude, by Lemma 11.3, that
\int_{Ω}{(f + g)^{+}} dm < ∞, \int_{Ω}{(f + g)^{−}} dm < ∞,
and this implies that f + g ∈ \mathcal{L^{1}}(Ω). To prove (11.14), observe that
f + g = (f + g)^{+} − (f + g)^{−}
and
f + g = f^{+} − f^{−} + g^{+} − g^{−},
hence
(f + g)^{+} + f^{−} + g^{−} = (f + g)^{−} + f^{+} + g^{+}
⇒\int_{Ω}{(f + g)^{+}} dm + \int_{Ω}{f ^{−}} dm + \int_{Ω}{g ^{−}} dm = \int_{Ω}{(f + g)^{−}} dm + \int_{Ω}{f ^{+}} dm + \int_{Ω}{g ^{+}} dm,
and we arrive at (11.14) by rearranging terms.
(v) If c ≥ 0 then (cf )^{+} = cf^{+} and (cf )^{−} = cf ^{−}, and if c < 0 then (cf )^{+} = −cf^{−} and (cf )^{−} = −cf ^{−}. In either case we arrive at Equation (11.15) by applying Lemma 11.3.
(vi) We can write f = g + f − g, which implies
\int_{Ω}{f } dm = \int_{Ω}{g } dm + \int_{Ω}{(f − g) } dm.
Since (f − g) ∈ \mathcal{L}_{+}^{0}, it follows that \int_{Ω}{(f − g) } dm ≥ 0 by definition.
(vii) Let A = {x ∈ Ω : f(x) > 0} and A_{n} = \left\{x ∈ Ω : f (x) > 1/n\right\}. (A_{n}) is an increasing sequence in \mathcal{M}, and from the theorem of Archimedes we see that it increases to A. Since m is continuous from below (Theorem 10.9),
m(A) = \underset{n→∞}{\lim} m(A_{n}).
For the purpose of evaluating \int_{Ω}{f } dm we may assume that f ≥ 0 on Ω, since the subset of Ω on which f is negative has measure 0. Thus
\int_{Ω}{f } dm = \int_{A_{n}}{f } dm + \int_{Ω \setminus A_{n}}{f } dm
≥ \int_{A_{n}}{f } dm
≥ \frac{1}{n} m(A{n}).
But \int_{Ω}{f } dm = 0, hence m(A_{n}) = 0 for every n, and therefore m(A) = 0.
(viii) Let E = {x ∈ Ω : f (x) ≥ 0}. We then have
\int_{Ω}{|f |} dm = \int_{E}{| f|} dm + \int_{Ω \setminus E}{|f |} dm
= \int_{E}{f} dm + \int_{Ω \setminus E}{(−f )} dm
= \int_{E}{f} dm − \int_{Ω \setminus E}{f } dm
= 0,
since both E and Ω\E are measurable. From (vii) we now see that |f|, and hence f, equals 0 almost everywhere.