[proofplan] We prove the four properties of the [Riemann integral](/page/Riemann%20Integral) in sequence. Linearity is established in two parts: scalar multiples (by case analysis on the sign of $\lambda$) and sums (by bounding the oscillation of $f + g$ by the sum of oscillations and using a common refinement). Monotonicity follows from linearity and non-negativity of the integral of a non-negative function. The absolute value property uses the [reverse triangle inequality](/theorems/2300) to control the oscillation of $|f|$, then applies monotonicity. Product integrability uses a telescoping identity and the boundedness of integrable functions. [/proofplan] [step:Prove integrability and linearity of the scalar multiple $\lambda f$] Let $f$ be Riemann integrable on $[a,b]$ and $\lambda \in \mathbb{R}$. **Case $\lambda \geq 0$:** For any subinterval $[x_{j-1}, x_j]$ of a partition $\mathcal{D}$, the identities $\sup_{[x_{j-1},x_j]} (\lambda f) = \lambda \sup_{[x_{j-1},x_j]} f$ and $\inf_{[x_{j-1},x_j]} (\lambda f) = \lambda \inf_{[x_{j-1},x_j]} f$ hold. Therefore the upper and lower Darboux sums satisfy $S(\lambda f, \mathcal{D}) = \lambda S(f, \mathcal{D})$ and $s(\lambda f, \mathcal{D}) = \lambda s(f, \mathcal{D})$, giving \begin{align*} S(\lambda f, \mathcal{D}) - s(\lambda f, \mathcal{D}) = \lambda(S(f, \mathcal{D}) - s(f, \mathcal{D})). \end{align*} Given $\epsilon > 0$, the [Integrability Criterion](/theorems/192) applied to $f$ provides a partition $\mathcal{D}$ 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: $\int_a^b \lambda f \, d\mathcal{L}^1 = \lambda \int_a^b f \, d\mathcal{L}^1$. **Case $\lambda < 0$:** Write $\lambda = -|\lambda|$. On each subinterval, $\sup(-|\lambda| f) = -|\lambda| \inf f$ and $\inf(-|\lambda| f) = -|\lambda| \sup f$ (negation reverses the ordering). Therefore \begin{align*} S(-|\lambda| f, \mathcal{D}) - s(-|\lambda| f, \mathcal{D}) = |\lambda|(S(f, \mathcal{D}) - s(f, \mathcal{D})). \end{align*} The same $\epsilon$-argument shows $\lambda f$ is integrable. For the integral value, the upper and lower Darboux integrals swap when multiplied by $-1$: $\overline{\int} (-f) \, d\mathcal{L}^1 = -\underline{\int} f \, d\mathcal{L}^1$, so $\int_a^b \lambda f \, d\mathcal{L}^1 = \lambda \int_a^b f \, d\mathcal{L}^1$. [/step] [step:Prove integrability and linearity of the sum $f + g$] Let $f, g$ be Riemann integrable on $[a,b]$. On each subinterval $[x_{j-1}, x_j]$ of a partition $\mathcal{D}$: \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*} Subtracting and summing over all subintervals: \begin{align*} S(f+g, \mathcal{D}) - s(f+g, \mathcal{D}) \leq \bigl(S(f, \mathcal{D}) - s(f, \mathcal{D})\bigr) + \bigl(S(g, \mathcal{D}) - s(g, \mathcal{D})\bigr). \end{align*} Given $\epsilon > 0$, choose partitions $\mathcal{D}_1$ and $\mathcal{D}_2$ with $S(f, \mathcal{D}_1) - s(f, \mathcal{D}_1) < \epsilon/2$ and $S(g, \mathcal{D}_2) - s(g, \mathcal{D}_2) < \epsilon/2$. Let $\mathcal{D} = \mathcal{D}_1 \cup \mathcal{D}_2$ be the common refinement. By the [Refinement Lemma](/theorems/193), refining a partition decreases upper sums and increases lower sums, so the bounds carry over: $S(f, \mathcal{D}) - s(f, \mathcal{D}) < \epsilon/2$ and $S(g, \mathcal{D}) - s(g, \mathcal{D}) < \epsilon/2$. Therefore $S(f+g, \mathcal{D}) - s(f+g, \mathcal{D}) < \epsilon$, and $f + g$ is integrable by the Integrability Criterion. For the integral value, the lower sum bound $s(f, \mathcal{D}) + s(g, \mathcal{D}) \leq s(f+g, \mathcal{D})$ and upper sum bound $S(f+g, \mathcal{D}) \leq S(f, \mathcal{D}) + S(g, \mathcal{D})$ yield, for any $\epsilon > 0$ and a sufficiently fine common partition $\mathcal{D}$: \begin{align*} \int_a^b f \, d\mathcal{L}^1 + \int_a^b g \, d\mathcal{L}^1 &\leq S(f, \mathcal{D}) + S(g, \mathcal{D}) \leq s(f+g, \mathcal{D}) + \epsilon + \epsilon \leq \int_a^b (f+g) \, d\mathcal{L}^1 + 2\epsilon, \end{align*} and similarly $\int_a^b (f+g) \, d\mathcal{L}^1 \leq \int_a^b f \, d\mathcal{L}^1 + \int_a^b g \, d\mathcal{L}^1 + 2\epsilon$. Since $\epsilon > 0$ was arbitrary, $\int_a^b (f+g) \, d\mathcal{L}^1 = \int_a^b f \, d\mathcal{L}^1 + \int_a^b g \, d\mathcal{L}^1$. [guided] The difficulty with sums is that the supremum of a sum can be strictly less than the sum of the suprema -- equality holds only when both functions attain their extrema at the same point. On each subinterval $[x_{j-1}, x_j]$: \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*} Subtracting and summing over all subintervals gives the oscillation bound: \begin{align*} S(f+g, \mathcal{D}) - s(f+g, \mathcal{D}) \leq \bigl(S(f, \mathcal{D}) - s(f, \mathcal{D})\bigr) + \bigl(S(g, \mathcal{D}) - s(g, \mathcal{D})\bigr). \end{align*} The common refinement $\mathcal{D} = \mathcal{D}_1 \cup \mathcal{D}_2$ is essential because the partitions making $f$ and $g$ individually well-approximated may be different. The [Refinement Lemma](/theorems/193) guarantees that passing to a finer partition only improves the approximation (upper sums decrease, lower sums increase), so a single partition works for both simultaneously. Choosing $\mathcal{D}_1$ with $S(f, \mathcal{D}_1) - s(f, \mathcal{D}_1) < \epsilon/2$ and $\mathcal{D}_2$ with $S(g, \mathcal{D}_2) - s(g, \mathcal{D}_2) < \epsilon/2$, the common refinement $\mathcal{D}$ inherits both bounds, giving $S(f+g, \mathcal{D}) - s(f+g, \mathcal{D}) < \epsilon$. For the integral identity, we squeeze $\int (f + g)\,d\mathcal{L}^1$ between $\int f\,d\mathcal{L}^1 + \int g\,d\mathcal{L}^1 - 2\epsilon$ and $\int f\,d\mathcal{L}^1 + \int g\,d\mathcal{L}^1 + 2\epsilon$ for every $\epsilon > 0$, forcing equality. [/guided] [/step] [step:Prove monotonicity of the integral] Suppose $f(x) \leq g(x)$ for all $x \in [a,b]$. Define $h := g - f$, so $h(x) \geq 0$ on $[a,b]$. By the linearity established above, $h$ is integrable. For any partition $\mathcal{D}$, the lower Darboux sum satisfies \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*} since $h(x) \geq 0$ implies $\inf_{[x_{j-1},x_j]} h \geq 0$ on each subinterval. Therefore \begin{align*} \int_a^b h \, d\mathcal{L}^1 = \sup_{\mathcal{D}} s(h, \mathcal{D}) \geq 0. \end{align*} By linearity: $\int_a^b g \, d\mathcal{L}^1 - \int_a^b f \, d\mathcal{L}^1 = \int_a^b (g - f) \, d\mathcal{L}^1 = \int_a^b h \, d\mathcal{L}^1 \geq 0$, giving $\int_a^b f \, d\mathcal{L}^1 \leq \int_a^b g \, d\mathcal{L}^1$. [/step] [step:Prove $|f|$ is integrable and establish the integral triangle inequality] For any $x, y$ in a subinterval $[x_{j-1}, x_j]$, the reverse triangle inequality gives $\bigl||f(x)| - |f(y)|\bigr| \leq |f(x) - f(y)|$. Taking the supremum over all pairs $x, y \in [x_{j-1}, x_j]$: \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*} The last equality holds because $\sup_{x,y} |f(x) - f(y)| = \sup f - \inf f$ on any bounded set. Summing over all subintervals: \begin{align*} S(|f|, \mathcal{D}) - s(|f|, \mathcal{D}) \leq S(f, \mathcal{D}) - s(f, \mathcal{D}). \end{align*} Since $f$ is integrable, for any $\epsilon > 0$ 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](/theorems/192), $|f|$ is integrable. For the triangle inequality, the pointwise bounds $-|f(x)| \leq f(x) \leq |f(x)|$ and the monotonicity property give \begin{align*} -\int_a^b |f| \, d\mathcal{L}^1 \leq \int_a^b f \, d\mathcal{L}^1 \leq \int_a^b |f| \, d\mathcal{L}^1, \end{align*} which is $\left|\int_a^b f \, d\mathcal{L}^1\right| \leq \int_a^b |f| \, d\mathcal{L}^1$. [/step] [step:Prove the product $fg$ is integrable] Since $f$ and $g$ are integrable on $[a,b]$, they are bounded: there exists $M > 0$ with $|f(x)| \leq M$ and $|g(x)| \leq M$ for all $x \in [a,b]$. For any $x, y \in [x_{j-1}, x_j]$, the identity $f(x)g(x) - f(y)g(y) = f(x)(g(x) - g(y)) + g(y)(f(x) - f(y))$ and the triangle inequality give \begin{align*} |f(x)g(x) - f(y)g(y)| &\leq |f(x)| \cdot |g(x) - g(y)| + |g(y)| \cdot |f(x) - f(y)| \\ &\leq M|g(x) - g(y)| + M|f(x) - f(y)|. \end{align*} Taking the supremum over $x, y \in [x_{j-1}, x_j]$ and summing: \begin{align*} S(fg, \mathcal{D}) - s(fg, \mathcal{D}) \leq M\bigl(S(g, \mathcal{D}) - s(g, \mathcal{D})\bigr) + M\bigl(S(f, \mathcal{D}) - s(f, \mathcal{D})\bigr). \end{align*} Given $\epsilon > 0$, choose a common refinement $\mathcal{D}$ with $S(f, \mathcal{D}) - s(f, \mathcal{D}) < \epsilon/(2M)$ and $S(g, \mathcal{D}) - s(g, \mathcal{D}) < \epsilon/(2M)$. Then $S(fg, \mathcal{D}) - s(fg, \mathcal{D}) < \epsilon$. By the [Integrability Criterion](/theorems/192), $fg$ is integrable. [guided] The product of two integrable functions need not satisfy any simple Darboux sum identity (unlike scalar multiples), so we use an indirect approach. The key algebraic identity is the "add-and-subtract" trick: \begin{align*} f(x)g(x) - f(y)g(y) = f(x)\bigl(g(x) - g(y)\bigr) + g(y)\bigl(f(x) - f(y)\bigr). \end{align*} This decomposes the oscillation of the product into two terms, each involving the oscillation of one factor multiplied by the bound on the other. The uniform bound $M$ on both $|f|$ and $|g|$ (which exists because Riemann integrable functions on a closed bounded interval are bounded) then converts the oscillation estimate into \begin{align*} \omega(fg, [x_{j-1}, x_j]) \leq M \cdot \omega(g, [x_{j-1}, x_j]) + M \cdot \omega(f, [x_{j-1}, x_j]), \end{align*} where $\omega(h, J) := \sup_J h - \inf_J h$ denotes the oscillation of $h$ on $J$. Summing over all subintervals and using the integrability of $f$ and $g$ to make both oscillation sums small completes the argument. [/guided] [/step]