Theorems Integral Properties Attributions

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Proof

**Step 1: Linearity — scalar multiple**

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Let $f$ be integrable on $[a,b]$ and $\lambda \in \mathbb{R}$. We show $\lambda f$ is integrable and $\int_a^b \lambda f = \lambda \int_a^b f$.

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**Case $\lambda \geq 0$:** For any subinterval $[x_{j-1}, x_j]$, $\sup_{[x_{j-1},x_j]} (\lambda f) = \lambda \sup_{[x_{j-1},x_j]} f$ and $\inf_{[x_{j-1},x_j]} (\lambda f) = \lambda \inf_{[x_{j-1},x_j]} f$. Therefore $S(\lambda f, \mathcal{D}) = \lambda S(f, \mathcal{D})$ and $s(\lambda f, \mathcal{D}) = \lambda s(f, \mathcal{D})$, so $S(\lambda f, \mathcal{D}) - s(\lambda f, \mathcal{D}) = \lambda(S(f, \mathcal{D}) - s(f, \mathcal{D}))$. Given $\epsilon > 0$, choose $\mathcal{D}$ with $S(f, \mathcal{D}) - s(f, \mathcal{D}) < \epsilon / (\lambda + 1)$. Then $S(\lambda f, \mathcal{D}) - s(\lambda f, \mathcal{D}) < \epsilon$, so $\lambda f$ is integrable. Taking the infimum over all upper sums: $\int_a^b \lambda f = \lambda \int_a^b f$.

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**Case $\lambda < 0$:** Write $\lambda = -|\lambda|$. On each subinterval, $\sup(-|\lambda| f) = -|\lambda| \inf f$ and $\inf(-|\lambda| f) = -|\lambda| \sup f$. Therefore $S(-|\lambda| f, \mathcal{D}) - s(-|\lambda| f, \mathcal{D}) = |\lambda|(S(f, \mathcal{D}) - s(f, \mathcal{D}))$. The same argument as above shows $\lambda f$ is integrable. The integral satisfies $\int_a^b \lambda f = \lambda \int_a^b f$ because the upper and lower Darboux integrals swap when multiplied by $-1$: $\overline{\int} (-f) = -\underline{\int} f$ and $\underline{\int} (-f) = -\overline{\int} f$.

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**Step 2: Linearity — sum**

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Let $f, g$ be integrable on $[a,b]$. On each subinterval $[x_{j-1}, x_j]$:

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\begin{align*} \sup_{[x_{j-1},x_j]} (f + g) &\leq \sup_{[x_{j-1},x_j]} f + \sup_{[x_{j-1},x_j]} g \\ \inf_{[x_{j-1},x_j]} (f + g) &\geq \inf_{[x_{j-1},x_j]} f + \inf_{[x_{j-1},x_j]} g \end{align*}

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Subtracting:

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\begin{align*} \sup(f+g) - \inf(f+g) \leq (\sup f - \inf f) + (\sup g - \inf g) \end{align*}

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Therefore $S(f+g, \mathcal{D}) - s(f+g, \mathcal{D}) \leq (S(f, \mathcal{D}) - s(f, \mathcal{D})) + (S(g, \mathcal{D}) - s(g, \mathcal{D}))$.

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Given $\epsilon > 0$, choose $\mathcal{D}_1$ with $S(f, \mathcal{D}_1) - s(f, \mathcal{D}_1) < \epsilon/2$ and $\mathcal{D}_2$ with $S(g, \mathcal{D}_2) - s(g, \mathcal{D}_2) < \epsilon/2$. Let $\mathcal{D} = \mathcal{D}_1 \cup \mathcal{D}_2$ (common refinement). By the Refinement Lemma, refining a partition decreases upper sums and increases lower sums, so $S(f, \mathcal{D}) - s(f, \mathcal{D}) < \epsilon/2$ and $S(g, \mathcal{D}) - s(g, \mathcal{D}) < \epsilon/2$. Therefore:

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\begin{align*} S(f+g, \mathcal{D}) - s(f+g, \mathcal{D}) < \epsilon \end{align*}

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By the Integrability Criterion, $f + g$ is integrable.

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For the integral value: since $s(f, \mathcal{D}) + s(g, \mathcal{D}) \leq s(f+g, \mathcal{D}) \leq S(f+g, \mathcal{D}) \leq S(f, \mathcal{D}) + S(g, \mathcal{D})$, taking the supremum over partitions on the left and infimum on the right: $\int_a^b f + \int_a^b g \leq \overline{\int_a^b} (f+g)$ and $\underline{\int_a^b} (f+g) \leq \int_a^b f + \int_a^b g$...

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More directly: for any partition $\mathcal{D}$, $s(f, \mathcal{D}) + s(g, \mathcal{D}) \leq s(f+g, \mathcal{D})$ and $S(f+g, \mathcal{D}) \leq S(f, \mathcal{D}) + S(g, \mathcal{D})$. Given $\epsilon > 0$, choose $\mathcal{D}$ fine enough that $S(f, \mathcal{D}) - s(f, \mathcal{D}) < \epsilon$, $S(g, \mathcal{D}) - s(g, \mathcal{D}) < \epsilon$, and $S(f+g, \mathcal{D}) - s(f+g, \mathcal{D}) < \epsilon$. Then:

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\begin{align*} \int_a^b f + \int_a^b g &\leq S(f, \mathcal{D}) + S(g, \mathcal{D}) \leq s(f, \mathcal{D}) + \epsilon + s(g, \mathcal{D}) + \epsilon \\ &\leq s(f+g, \mathcal{D}) + 2\epsilon \leq \int_a^b (f+g) + 2\epsilon \end{align*}

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Similarly $\int_a^b (f+g) \leq S(f+g, \mathcal{D}) \leq S(f, \mathcal{D}) + S(g, \mathcal{D}) \leq \int_a^b f + \epsilon + \int_a^b g + \epsilon$. So $|\int_a^b (f+g) - \int_a^b f - \int_a^b g| \leq 2\epsilon$. Since $\epsilon > 0$ was arbitrary, $\int_a^b (f+g) = \int_a^b f + \int_a^b g$.

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**Step 3: Monotonicity**

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If $f(x) \leq g(x)$ for all $x \in [a,b]$, then $h = g - f \geq 0$. By Step 2, $h$ is integrable. For any partition $\mathcal{D}$:

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\begin{align*} s(h, \mathcal{D}) = \sum_{j=1}^{n} \inf_{[x_{j-1},x_j]} h(x) \cdot (x_j - x_{j-1}) \geq 0 \end{align*}

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since $h(x) \geq 0$ implies $\inf_{[x_{j-1},x_j]} h \geq 0$. Therefore:

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\begin{align*} \int_a^b h = \sup_{\mathcal{D}} s(h, \mathcal{D}) \geq 0 \end{align*}

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By linearity: $\int_a^b g - \int_a^b f = \int_a^b (g - f) = \int_a^b h \geq 0$, giving $\int_a^b f \leq \int_a^b g$.

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**Step 4: Absolute value integrability and triangle inequality**

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For any $x, y \in [x_{j-1}, x_j]$, the reverse triangle inequality gives $||f(x)| - |f(y)|| \leq |f(x) - f(y)|$. Therefore:

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\begin{align*} \sup_{[x_{j-1},x_j]} |f| - \inf_{[x_{j-1},x_j]} |f| \leq \sup_{x,y \in [x_{j-1},x_j]} |f(x) - f(y)| = \sup_{[x_{j-1},x_j]} f - \inf_{[x_{j-1},x_j]} f \end{align*}

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(The last equality holds because $\sup_{x,y} |f(x) - f(y)| = \sup f - \inf f$ on any set for a bounded function.) Summing over all subintervals:

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\begin{align*} S(|f|, \mathcal{D}) - s(|f|, \mathcal{D}) \leq S(f, \mathcal{D}) - s(f, \mathcal{D}) \end{align*}

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Since $f$ is integrable, given $\epsilon > 0$ there exists $\mathcal{D}$ with $S(f, \mathcal{D}) - s(f, \mathcal{D}) < \epsilon$, hence $S(|f|, \mathcal{D}) - s(|f|, \mathcal{D}) < \epsilon$. By the Integrability Criterion, $|f|$ is integrable.

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Since $-|f(x)| \leq f(x) \leq |f(x)|$ for all $x$, monotonicity (Step 3) gives:

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\begin{align*} -\int_a^b |f| \leq \int_a^b f \leq \int_a^b |f| \end{align*}

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which is $\left|\int_a^b f\right| \leq \int_a^b |f|$.

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**Step 5: Product integrability**

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Since $f$ and $g$ are integrable on $[a,b]$, they are bounded: $|f(x)| \leq M$ and $|g(x)| \leq M$ for some $M > 0$ and all $x \in [a,b]$. For any $x, y \in [x_{j-1}, x_j]$:

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\begin{align*} |f(x)g(x) - f(y)g(y)| &= |f(x)g(x) - f(x)g(y) + f(x)g(y) - f(y)g(y)| \\ &\leq |f(x)||g(x) - g(y)| + |g(y)||f(x) - f(y)| \\ &\leq M|g(x) - g(y)| + M|f(x) - f(y)| \end{align*}

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Taking the supremum over $x, y \in [x_{j-1}, x_j]$:

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\begin{align*} \sup(fg) - \inf(fg) \leq M(\sup g - \inf g) + M(\sup f - \inf f) \end{align*}

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Summing: $S(fg, \mathcal{D}) - s(fg, \mathcal{D}) \leq M(S(g, \mathcal{D}) - s(g, \mathcal{D})) + M(S(f, \mathcal{D}) - s(f, \mathcal{D}))$.

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Given $\epsilon > 0$, choose $\mathcal{D}$ with $S(f, \mathcal{D}) - s(f, \mathcal{D}) < \epsilon/(2M)$ and $S(g, \mathcal{D}) - s(g, \mathcal{D}) < \epsilon/(2M)$ (using a common refinement as in Step 2). Then $S(fg, \mathcal{D}) - s(fg, \mathcal{D}) < \epsilon$. By the Integrability Criterion, $fg$ is integrable.

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