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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.

A positive linear functional on the space of continuous compactly supported functions of a locally compact Hausdorff space has a unique representing positive Radon measure. Positivity gives boundedness on fixed compact supports: choose a compactly supported continuous cutoff $p\ge 1$ on the support $K$, then $|I(f)|\le \Vert f{\Vert }_{\infty }I(p)$ for real $f$ supported in $K$. The resulting measure is finite on compact sets but need not have finite total mass.