Is Zero Outer Content Sufficient for Function Integrability?

[STEMucator]

Homework Statement

1. Suppose that $f = 0$ at all points of a rectangle $R$ except on a set $D$ of outer content zero, where $f \geq 0$. If $f$ is bounded, prove that $f$ is integrable on $R$ and $\int \int f dA = 0$.

2. Now suppose $f$ is an integrable function on a rectangle $R$ and let $g$ be bounded on $R$. If $f = g$ except for $(x,y)$ in some set of outer content zero where $f ≤ g$, prove that $g$ is integrable on $R$ and $\int \int f dA = \int \int g dA$.

Homework Equations

The Attempt at a Solution

1. Suppose $f ≤ M$ on $R$. For all $\epsilon > 0$, we can choose a partition $P$ such that the area enclosed by the partition satisfies:

$$A_P < \frac{\epsilon}{M}$$.

Then we have:

$$S_P = \sum_{i} M_i \Delta A_i ≤ M A_P < \epsilon$$

This is because $M_i = 0$ for rectangles which do not contribute to $A_P$. Then $\inf(S_P) < \epsilon \Rightarrow \inf(S_P) \leq 0$.

We know that $\sup(s_p) \geq 0$ and so $0 \leq sup(s_p) \leq inf(S_P) \leq 0$. So obviously $sup(s_p) = inf(S_P)$.

Hence it must be the case that $f$ is integrable since the upper and lower sums converge, and $\int \int f dA = 0$.

I think that should be it for the first question hopefully.

2. Well we have $g - f = 0$ on $R$ except on a set of outer content zero where we have $g - f \geq 0$. So by the first question, $g - f$ is integrable.

Hence $g = (g - f) + f$ is integrable by question 1. So we can finally write:

$$\int \int f dA = \int \int (g - f) dA + \int \int f dA = \int \int g dA$$

Need a sanity check to make sure this isn't wrong. Been a long while since I had to think about a proof that hard.