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There is an essential qualification to the printed hypothesis. If compactly generated group means merely generated algebraically by a compact subset, the claim is false. In the additive group $\mathbb{Q}$ with its usual topology,is compact, being a convergent sequence together with its limit, and generates every rational number since any integer denominator divides some factorial. The group is metrizable. However, an invariant Radon measure has a common mass $c$ on each singleton. If $c>0$, the compact set $K$ has infinite measure, contradicting finiteness on compact sets. If $c=0$, countable additivity makes the entire countable group have measure zero. Thus it has no nonzero Haar measure. This proves that compact generation does not imply local compactness and disproves the literal unrestricted statement.

The intended existence statement is valid when local compactness is included, either explicitly or in the convention that the group has a compact generating neighbourhood. Here is a construction for a locally compact, compactly generated, metrizable Hausdorff group. No invariant measure is assumed during its construction.

Enlarge a compact generating set to a symmetric compact neighbourhood of the identity element $S$. Then $G={\bigcup }_n{S}^n$, and each $S^n$ is compact. Moreover $S^n$ lies in the interior of $S^{n+1}$, since multiplication by the open neighbourhood of the identity element contained in $S$ supplies an open neighbourhood of every point of $S^n$. This gives a compact exhaustion. Every compact metric space has a countable dense subset: take finite $1/k$-nets and unite them. Their union over this exhaustion is dense in $G$, so the metric space is separable and has a countable open basis.

We use nonnegative compactly supported continuous functions and cutoffs equal to one on specified compact sets. Their existence is covered by the source's permission concerning auxiliary functions. Fix a nonzero such function $h$. For nonzero $u\ge 0$ define the weighted covering quantityThe quantities are finite: an open set where $u$ is bounded below has translates covering the compact support of $f$, so finitely many sufficiently large weighted translates dominate it. The denominator is positive, since evaluation at a point where $h>0$ gives $(h:u)\ge h(y)/\Vert u{\Vert }_{\infty }>0$.

Each $I_u$ is nonnegative, homogeneous, monotone, left invariant and subadditive; these follow respectively by rescaling, comparing or combining covers, and translating their centers. Also $I_u(h)=1$. Composing a cover of $f$ by translates of $h$ with a cover of $h$ by translates of $u$ givesThis is a bound independent of $u$. On functions supported in a fixed compact set $C$, choose a cutoff $k\ge 1$ on $C$. Monotonicity and subadditivity implyThus the covering functionals are uniformly Lipschitz on each fixed support.

Choose a nonzero $u_n\ge 0$ with support in an neighbourhood of the identity element shrinking to $e$. The nonnegative functions supported in each compact member of the exhaustion form a separable metric cone in the supremum norm, as a subspace of continuous functions on that compact metric space. Select a countable dense family from each cone and include $h$. The uniform scalar bounds permit a diagonal subsequence along which $I_{u_n}$ converges on their countable union. The displayed Lipschitz estimate extends convergence to every nonnegative member of $C_c(G)$. Denote the limit by $I$. It inherits positivity, homogeneity, monotonicity, subadditivity, invariance and $I(h)=1$.

The crucial remaining step is additivity. Fix $f_1,f_2\ge 0$ with combined support $C$, and fix a compact neighbourhood of the identity element $V_0$. Choose a cutoff $k$ equal to one on $C{V}_0^{-1}{V}_0$. For $\delta >0$, putsetting the ratio to zero off the denominator's nonzero set. These ratios are continuous: $k=1$ on a neighbourhood of the supports, and off those supports the numerator vanishes. Their sum is at most one. Compactness and continuity give, for each $\eta >0$, an neighbourhood of the identity element $V$ small enough that each $r_i$ oscillates by at most $\eta$ on any translate of $V$ meeting $C$. To justify this uniform choice, such translates lie inside the fixed compact set $C{V}_0^{-1}{V}_0$ when $V\subset V_0$; continuity of $(y,v)\mapsto r_i(yv)$ and a finite cover of that compact set give a single small right-increment neighbourhood, then choose $V^{-1}V$ inside it.

If $u$ has support in $V$ and $F\le {\sum }_j{c}_j{T}_{x_j}u$, discard translates not meeting $C$ when covering $f_i$. Let $a_{ij}$ be the supremum of $r_i$ on the remaining translate's support. Then $f_i\le {\sum }_j{a}_{ij}{c}_j{T}_{x_j}u$, and comparing both suprema with the ratios at one common point gives $a_{1j}+a_{2j}\le 1+2\eta$. Taking covering infima yieldsAlong the subsequence this holds eventually, since its supports shrink. Pass to $I$, then let $\eta \downarrow 0$ and $\delta \downarrow 0$; the uniform bound on $I_u(k)$ gives $I(f_1)+I(f_2)\le I(f_1+f_2)$. Subadditivity gives the reverse inequality. Thus $I$ is additive.

Extend $I$ linearly to real functions using positive and negative parts, and then complex-linearly to $C_c(G)$. The compact-support bounds above make it a positive, locally bounded functional. The Riesz representation on compactly supported continuous functions gives a Radon measure $\mu$ with $I(f)=\int fd\mu$. Translation invariance of $I$ and uniqueness in that representation give $\mu (xE)=\mu (E)$ for Borel $E$. It is nonzero because $I(h)=1$, and finite on compact sets because a cutoff dominating their indicators has finite integral. Finally every nonempty open set has positive measure: otherwise its translates cover the compact support of $h$, a finite subcover would give that support measure zero, contradicting $I(h)=1$. Hence $\mu$ is a Haar measure under the stated local-compactness qualification. This is Haar measure from normalized covering functionals, with the limiting additivity proved explicitly.

There is a definitional qualification. If a compactly generated group means a group algebraically generated by a compact subset, the assertion as printed is false. In the usual topology on the additive group $\mathbb{Q}$, the setis compact and generates $\mathbb{Q}$: every positive integer denominator divides some factorial. The group is metrizable, but it has no nonzero Haar measure finite on compact sets. Indeed, translation invariance gives a common singleton mass $a\ge 0$. If $a>0$, the infinite compact set $K$ has infinite mass; if $a=0$, countable additivity gives zero mass to the whole countable group. This proves that compact generation does not imply local compactness or existence of Haar measure.

If the convention includes a compact generating neighbourhood, local compactness is already part of the hypothesis. More explicitly, the standard corrected result is every locally compact, compactly generated metrizable group has Haar measure. Here is a construction, rather than an invocation of that existence theorem.

Fix a nonzero $h\ge 0$ in the space of continuous compactly supported functions $C_c(G)$. For nonzero $u\ge 0$ in $C_c(G)$ and $f\ge 0$ in $C_c(G)$, define the covering number and its normalized functional byThese numbers are finite: translates of the region where $u>0$ cover the compact support of $f$, and a finite subcover, multiplied by a sufficiently large coefficient, dominates $f$. They are positive when $f\ne 0$, since every cover has total coefficient at least $\Vert f{\Vert }_{\infty }/\Vert u{\Vert }_{\infty }$. Composition of covers gives $(f:u)\le (f:h)(h:u)$. Thus, with $M_f=(f:h)$,The functionals are homogeneous, monotone, subadditive and exactly invariant under left translation; the last property follows by relabelling the covering translates.

The key step in Haar measure from normalized covering functionals is asymptotic additivity as $\mathrm{supp}u$ shrinks to $e$. Given $f_1,f_2\ge 0$, put $f=f_1+f_2$ and choose $p\in C_c(G)$ with $p\ge 1$ on both supports. For $\delta >0$, set ${\alpha }_i=f_i/(f+\delta p)$, extended by zero where the denominator vanishes. These are continuous compactly supported functions and ${\alpha }_1+{\alpha }_2\le 1$. For any $\eta >0$, a sufficiently small identity neighbourhood $V$ ensuresTake $\mathrm{supp}u\subseteq V$ and any cover $f\le {\sum }_j{a}_j{T}_{x_j}u$, where $T_{x}u(t)=u(x^{-1}t)$. Since $f_i\le {\alpha }_{i}f+\delta p$, it follows thatAdding and then taking the infimum over the cover of $f$ givesFirst make $\delta$ small, then $\eta$ small, and finally the support of $u$ small enough for the oscillation bound. The additivity defect therefore tends to zero.

Compact generation gives a countable compact cover by finite products of a compact symmetric generating set. Local compactness upgrades this to a compact exhaustion with interiors covering $G$. Metrizability supplies a countable identity-neighbourhood base and a countable family of nonnegative $C_c$ functions dense in the uniform norm on each fixed compact support; this is the countable family whose existence may be assumed. Choose nonzero $u_n\ge 0$ with supports shrinking to $e$. The uniform bounds $I_{u_n}(f)\le M_f$ and a diagonal subsequence give convergence on that countable family. To extend convergence to every nonnegative $f$ supported in a fixed compact $L$, choose a cutoff $p_L\ge 1$ on $L$. Monotonicity and subadditivity givefor $f,g$ supported in $L$. Uniform approximation therefore makes the chosen subsequence Cauchy on every such $f$.

Let $I(f)$ be its limit. Asymptotic additivity, homogeneity and invariance give an additive positive invariant functional on the nonnegative cone, with $I(h)=1$. Extend it by differences to real $C_c(G)$ and then complex linearly. It is bounded on each fixed compact support by the cutoff bound. The Riesz representation on compactly supported continuous functions consequently produces a positive Radon measure $m$ with $I(f)=\int fdm$. Invariance of $I$ and uniqueness of that representation imply $m(xE)=m(E)$. The measure is nonzero because $I(h)=1$, and finite on compact sets by the cutoff estimate. Moreover, for any nonzero $f\ge 0$, composition of covers gives $I_u(f)\ge 1/(h:f)>0$, so every nonempty open set has positive measure. HenceThis completes the intended construction under the convention supplying local compactness, while the rational example settles the weaker literal interpretation.