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I'm very new to this material and I would like someone more experienced to give input if possible.

$Q \subset \mathbb R^n$ is a box and $f: Q \to \mathbb R$ is a function. Let $\Xi_Q$ be the set of all sub-boxes of $Q$, and $F: \Xi_Q \to \mathbb R$ be another function.

Suppose the following 2 statements hold:

1) For any $B \in \Xi_Q$ we have: $v(B)\cdot \inf_{x \in B}f(x) \leq F(B) \leq v(B) \cdot \sup_{x \in B}f(B)$.

2) For any $B_1,B_2, \dots, B_k \in \Xi_Q$ which are disjoint in pairs we have: $F(B_1\cup \dots \cup B_k) = \sum_{i=1}^{k}F(B_i)$

Show that $$\int_{\overline{B}} f \leq F(B)\leq\int^{\overline{}}_{B}f$$ For all $B \in \Xi_Q$.

Clarification: $\int_{\overline{B}}f = \sup_{\Pi}\underline{S}(f,\Pi)$ and $\int_{B}^{\overline{}}f=\inf_{\Pi}\overline{S}(f,\Pi)$ where $\Pi$ is a partition of the box $B$ and $\underline{S},\overline{S}$ are lower and upper Darboux sums respectively.

What I did:

$$\int_{\overline{B}}f = \sup_{\Pi}\underline{S}(f,\Pi) = \sup_\Pi\{\sum_{R \in \Pi}\inf_{x \in R}f(x)\cdot v(R)\} \leq \sup_\Pi\{\sum_{R \in \Pi}F(R)\} = \sup_\Pi\{F(B)\} = F(B)$$

The inequality comes from statement number $1$, as $R \in \Xi_Q$ if I understand correctly. The equality after it comes from the fact that $\Pi$ is a partition of $B$, so all $R \in \Pi$ are disjoint, and their union is $B$, so from statement number $2$ the equality follows.

Now for the other direction, a very similar idea:

$$\int^{\overline{}}_{B}f = \inf_{\Pi}\overline{S}(f,\Pi) =\inf_\Pi\{\sum_{R \in \Pi}\sup_{x \in R}f(x)\cdot v(R)\} \geq \inf_\Pi\{\sum_{R \in \Pi}F(R)\} = \inf_\Pi\{F(B)\} = F(B)$$

Which proves what we wanted to show.

Is this the correct method? I know it's simple but I'm very new to this material and I missed the lecture sadly.

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