Proving that $\int_A f = \underline \int_Q f_A$ for a non-negative function defined on an open subset $A$ of $\mathbb{R}^n$

Question

In the book of Analysis On Manifolds by Munkres, at page 133, question 7, it is asked that

Proof:

We only show that for a non-negative $f$, we have

$$\int_A f = \underline \int_{Q} (f)_A ,$$ where $\int_A f$ denotes the improper integral, whereas in the RHS, we have a the definite integral.

Since $A$ is bounded, it has been shown in the book that $\int_A f$ exists, and we can always talk about the lower integral of a function.

Now, let $(D_n)$ be a sequence of compact rectifiable sets in $A$ s.t $D_{n+1} \subset D_n$ and $$\bigcup_n D_n = A.$$ [The existence of such a sequence of compact rectifiable sets is proved in the book.]

Observe that for any $n$,

$$\int_{D_n} f = \int_{D_n} f_A = \underline \int_{D_n} f_A = , \quad since \quad f = f_A \quad on \quad A.$$

But by monotonicity of definite integral, we also have

$$\underline \int_{D_n} f_A \leq \underline \int_Q f_A.$$

Since $n$ was arbitrary, by definition of improper integral, we have

$$\int_A f \leq \underline \int_Q f_A.$$

On the other hand,

Let $P$ be a partition of $Q$, and $R_1,..., R_k$ be subrectangles contained in $A$, and define $D = \bigcup_i^k R_i \subset A.$

Observe that

$$L(f_A, P) = L(f_A, D)$$ because if a subrectange has an non-empty intersection with both $Q-A$, then $m(f_A) = 0$ on that rectange, so the only contribution to the lower integral are going to come from those rectangles that are contained by $A$ completely.

Hence,

$$L(f_A, P) = L(f, D) = (ordinary) \int_D f \leq (extended) \int_A f$$ Note that since $D$ is compact rectifiable, $\int_D f$ in the ordinary sense exists.

But, now, since $P$ was arbitrary, we get $$\underline \int_Q f_A \leq \int_A f.$$

Hence the proof is complete.

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Questions:

First of all, is there any flaw in the proof ? I mean it took my almost a day to write completely complete this proof.

Secondly, is there any problem with especially about the argument $L(f_A, P) = L(f_A, D)$ ?

Answer 1

This is very close to being correct.

In the first part, the sequence $(D_n)$ needs the property $D_n \subset \text{Int } D_{n+1}$ (not $D_{n+1} \subset D_n)$ and such a sequence is shown to exist in Munkres.

In the second part, $D$ is a subset of $A$, not a partition, so the notation $L(f,D)$ is meaningless and does not represent a lower sum.

However, $L(f_A,P)$ has non-zero contributions only from subrectangles that form $D$ where $f_A = f$, so

$$L(f_A,P) = \sum_{i=1}^k \inf_{x \in R_i} f(x) \, vol(R_i) \leqslant \sum_{i=1}^k \int_{R_i} f = \int_Df \leqslant \int_Af,$$

where the last integral is extended (as you saw) and, therefore,

$$\underline{\int}_Q f_A = \sup_P L(f_A,P) \leqslant \int_A f$$