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For $x\ge 0$, the function $h(x)=e^x-1-x$ has $h(0)=0$ and $h^{\prime }(x)=e^x-1\ge 0$, because the exponential function is increasing and $e^0=1$. Therefore $\boxed{e^x\ge 1+x}$.

Write $S_n={\sum }_{j=1}^n{a}_j$ and $P_n={\prod }_{j=1}^n(1+a_j)$. Expanding the product produces $1+S_n$ and additional nonnegative terms, giving $P_n\ge 1+S_n\ge S_n$. Applying the exponential inequality to every factor givesBoth sequences are increasing, since the terms $a_j$ are positive. If $S_n$ converges to a finite value $S$, then $P_n\le e^S$, so the convergence theorem for monotone bounded sequences makes $P_n$ converge. Conversely, if $P_n$ has a finite limit, then $S_n\le P_n$ is bounded and increasing, hence converges by the same theorem. Thus the positive series and the product both converge to finite limits, or both diverge to $+\infty$. This proves the positive sum-product convergence criterion. The word limit here means a finite real limit; allowing extended infinite limits would make the conclusion vacuous for these monotone sequences.

For a partition $D=\{a=x_0<x_1<\cdots <x_n=b\}$, put $\Delta x_k=x_k-x_{k-1}$ andBoundedness makes these finite. The lower Darboux sum and upper Darboux sum areSince $m_k\le M_k$ and the widths are positive, $s(f,D)\le S(f,D)$.

Splitting a partition interval into smaller intervals can only increase each infimum and decrease each supremum. The smaller widths sum to the original width. Its new lower contribution is therefore at least its old lower contribution, and its new upper contribution at most its old upper contribution. Repeating this for every inserted point proves Darboux sum refinement monotonicity:For arbitrary partitions, take their common partition refinement $D_{\ast }=D_1\cup D_2$. ThenNo compatibility of the original partitions is required.

Finally suppose $f\ge 0$ is Riemann integrable. Then $t_k=m_k\Delta x_k\ge 0$, so every product factor is nonnegative and the exponential inequality can be multiplied safely:A lower Darboux sum is no greater than the Riemann integral, since the integral equals the supremum of all lower sums. Monotonicity of the exponential therefore gives the exponential bound for a lower-sum productFor a degenerate interval $a=b$, the empty product and the exponential are both one.

First apply the Cantor theorem by diagonalization. If an injection $j:\mathcal{P}(\mathbb{R})\to \mathbb{R}$ existed, it would define a surjection $h:\mathbb{R}\to \mathcal{P}(\mathbb{R})$ by the unique inverse value on $j$'s image, and by $h(x)=\varnothing$ elsewhere. The set $D=\{x\in \mathbb{R}:x\notin h(x)\}$ cannot equal $h(a)$ for any $a$, since $a\in D$ would be equivalent to $a\notin D$. This contradicts surjectivity. There is no injection from the power set of the real line into the real line.

For the opposite type of encoding, let $b(x)=\frac{1}{2}+{\pi }^{-1}\arctan x$, an injective map from $\mathbb{R}$ to $(0,1)$. Give $b(x)$ its canonical binary expansion, choosing the terminating expansion with trailing zeros whenever two expansions exist. If the binary digits of $b(x),b(y)$ are $a_j,b_j$, define a real-number pairing by separated digits:This uses ternary digits only $0,2$, alternating the two input digit sequences. If two outputs first differ at ternary position $k$, that difference has magnitude $2\cdot 3^{-k}$, while every later digit together contributes at most ${\sum }_{j>k}2\cdot 3^{-j}=3^{-k}$. Their outputs therefore differ. The digits recover both binary sequences, hence both inputs. Neither output endpoint $0$ nor $1$ occurs because the input numbers lie strictly between zero and one. Thus $H$ is an injection from ${\mathbb{R}}^2$ into $(0,1)\subset \mathbb{R}$. The ternary construction avoids an unjustified decimal-interleaving argument at ambiguous expansions.

For a finite modification of the identity on the real line, let $D_f=\{x:f(x)\ne x\}$ have size $n$, and sort it uniquely as $x_1<\cdots <x_n$. Its finite record is $(x_1,f(x_1),\dots ,x_n,f(x_n))$. It determines $f$ completely, with the identity used outside the recorded points. Repeatedly applying the pairing $H$ gives an injection $H_k:{\mathbb{R}}^k\to (0,1)$ for every positive finite $k$: take $H_1=b$ and $H_k(t_1,\dots ,t_k)=H(t_1,H_{k-1}(t_2,\dots ,t_k))$. Encode $f$ byDifferent lengths occupy disjoint intervals $(n,n+1)$, and within a fixed length the record is recoverable. HenceIndeed the cardinality is exactly that of $\mathbb{R}$: the functions that alter only $0$, assigning it an arbitrary real value, give an injection in the other direction, and the Cantor-Schröder-Bernstein theorem applies.