Question 11.T.8: If f ∈ L¹(Ω) then, for every ε > 0, there exists (i) a st...
If f ∈ \mathcal{L}^{1}(Ω) then, for every ε > 0, there exists
(i) a step function ψ on Ω and a compact interval I such that ψ(x) = 0 for all x ∉ I, and
\int_{Ω} |f − ψ| dm < ε,
(ii) a continuous function g on Ω and a compact interval I such that g(x) = 0 for all x ∉ I, and
\int_{Ω} |f − g| dm < ε.
(i) we shall prove this part in stages:
(a) To begin with, assume f = χ_{E} where E is a measurable subset of Ω. Since f ∈ \mathcal{L}^{1}(Ω), m(E) < ∞. By Theorem 10.15 there is a set F ∈ \mathcal{E} such that m(EΔF ) < ε. Let ψ = χ_{F} . Since F = \cup_{i=1}^{n} [a_{i}, b_{i}), where the intervals [a_{i}, b_{i}) are pairwise disjoint, ψ is a step function. Clearly, ψ = 0 outside [a, b], where a = \min\left\{a_{1}, · · · , a_{n}\right\} and b = \max\left\{b_{1}, · · · , b_{n}\right\}.
Furthermore,
\int_{Ω} |f − ψ| dm = \int_{Ω} |χ_{E} − χ_{F} | dm
= \int_{Ω}χ_{EΔF} dm
= m((EΔF ) ∩ Ω) < ε.
(b) Suppose f = \Sigma_{i=1}^{n} c_{i}χ_{E_{i}} ∈ \mathcal{S}_{+}(Ω). Let c = \max\left\{c_{1}, · · · , c_{n}\right\}, and, for each 1 ≤ i ≤ n, choose a step function ψ_{i} as in (a) such that
\int_{Ω} \left|χ_{E_{i}} − ψ_{i}\right| dm < \frac{ε}{nc} .
Now ψ = \Sigma_{i=1}^{n} c_{i}ψ_{i} is a step function which vanishes (i.e., equals 0) outside a compact interval and satisfies
\int_{Ω}|f − ψ| dm ≤ \sum\limits_{i=1}^{n} {c_{i} \int_{Ω} \left|χ_{E_{i}} − ψ_{i}\right| dm} < ε.
(c) Suppose now that f ∈ \mathcal{L}_{+}^{0}(Ω). By the definition of the integral, we can choose \varphi ∈ \mathcal{S}_{+}(Ω) such that φ ≤ f and 0 ≤ \int_{Ω} f dm− \int_{Ω} \varphi dm < ε/2.
Since \varphi ∈ \mathcal{L}^{1}(Ω), by part (b) there is a step function ψ which vanishes outside a compact interval and satisfies
\int_{Ω} |\varphi − ψ| dm < \frac{ε}{2} ,
and therefore
\int_{Ω} |f − ψ| dm ≤ \int_{Ω} |f − \varphi| dm + \int_{Ω} |\varphi − ψ| dm < ε.
(d) Finally, let f ∈ \mathcal{L}^{1}(Ω). According to part (c), there are two step functions ψ_{1} and ψ_{2}, both of which vanish outside a compact interval, such that
\int_{Ω} \left|f^{+} − ψ_{1} \right| dm < \frac{ε}{2} , \int_{Ω} \left|f^{−} − ψ_{2} \right| dm < \frac{ε}{2} .
Consequently, ψ = ψ_{1} − ψ_{2} is a step function which vanishes outside a compact interval and satisfies
\int_{Ω} |f − ψ| dm ≤ \int_{Ω} \left|f^{+} − ψ_{1} \right| dm + \int_{Ω} \left|f^{−} − ψ_{2} \right| dm < ε.
(ii) To prove the second part of the theorem, choose a step function ψ which vanishes outside a compact interval and satisfies
\int_{Ω} |f − ψ| dm < \frac{ε}{2} .
Suppose ψ = \Sigma_{i=1}^{n} c_{i}χ_{[a_{i},b_{i})}, where the intervals are pairwise disjoint.
Let 0 < ε^{\prime} < ε / n \max |c_{i}| , and define g_{i} : [a_{i}, b_{i}) → \mathbb{R} by
g_{i}(x) = \begin{cases} 2(x − a_{i})/ε^{\prime}, & x ∈ [a_{i}, a_{i} + ε^{\prime}/2) \\ 1, & x∈ [a_{i} + ε^{\prime}/2, b_{i} − ε^{\prime}/2) \\ −2(x − b_{i})/ε^{\prime}, & x ∈ [b_{i} − ε^{\prime}/2, b_{i}) \\ 0, & x \notin [a_{i}, b_{i}). \end{cases}
Each g_{i}, and hence g = \Sigma_{i=1}^{n} c_{i}g_{i}, is a continuous function which vanishes outside \cup_{i=1}^{n} [a_{i}, b_{i}). Furthermore,
\int_{Ω} |ψ − g| dm ≤ \sum\limits_{i=1}^{n} {|c_{i}| \int_{Ω} \left|χ_{[a_{i} , b_{i})} − g_{i}\right| dm}
= \sum\limits_{i=1}^{n}{|c_{i}| \frac{ε^{\prime}}{2} < \frac{ε}{2}} .
Hence \int_{Ω} |f − g| dm ≤ \int_{Ω} |f − ψ| dm + \int_{Ω} |ψ − g| dm < ε.
