For risk-averse people with many good alternatives for spending small sums of money, an occasional lottery play is portfolio diversification.
For poor people or ones without good alternative micro-investments (and, typically, many bad options), there are all sorts of reasons why saving one more coin is not necessarily more appealing than using it sometimes to purchase a lottery ticket.
Expected value is a meaningless metric for the lotteries with low odds and low entry costs. The positive part of the expectation would usually take thousands of lifetimes to realize, and the negative total can be mitigated or maybe even reversed (the analysis is complicated) by playing selectively when the jackpot is large.
One of the more famous Berkeley mathematicians (Chern?) had a Ph.D student who won millions of USD in a lottery and donated some of the money to the department. It is hard to say how many such windfalls might have been lost by math departments that dutifully taught students never to invest for negative expected returns, but it is food for thought.
I now have a condition which I believe is necessary and sufficient in the special case of a finite sample space. My attempted proof is below.
Claim
Let $\Omega = [n]$, equipped with probability $Q=(q_1,\ldots,q_n)$ in the probability simplex in $R^n$. For a given risk metric $\rho:R^n \rightarrow R$, a latent disutility function $f:R \rightarrow R$ exists satisfying $\rho(Z)=E(f \circ Z)$ if and only if for every $\{x_1, \ldots, x_n\} \subseteq R$, $${\scr Q{\bf f}=R},$$ where $\scr Q$ is an $\underbrace{n \times \cdots \times n}_\text{$n+1$ times}$ tensor, ${\bf f} \in R^n$, and $\scr R$ is an $\underbrace{n \times \cdots \times n}_\text{$n$ times}$ tensor, with the components
$${\scr q}_{\omega_1 \ldots \omega_n k}=\sum_j q_j I(\omega_j=k)$$
$${\bf f}_\omega = \rho(x_\omega, \ldots, x_\omega)$$
$${\scr r}_{\omega_1 \ldots \omega_n}=\rho(x_{\omega_1}, \ldots, x_{\omega_n}).$$.
Proof
$(\implies)$ Assume such an $f$ exists. A cost variable $Z=(z_1,\ldots,z_n)$ is an element of $R^n$, and $E(Z) =\sum_j q_j z_j$. If $Z$ is constant, then $z_j = c$ for each $j \in [n]$, and thus $$E(f \circ Z) =\sum_j q_jf(z_j)= \sum_j q_jf(c)=f(c).$$
Hence $f(c) = \rho(c,\ldots,c)$ for each $c \in R$.
For any $\{x_1, \ldots, x_n\} \subseteq R$, there are $n^n$ random variables supported on the $x_i$, each of which has the form $Z = (x_{\omega_1}, \ldots, x_{\omega_n})$. Each such $Z$ places a linear constraint on $\rho$: $$\rho(x_{\omega_1}, \ldots, x_{\omega_n})=\rho(Z)=E(f \circ Z)=\sum_j q_jf(x_{\omega_j})=\sum_j q_j\rho(x_{\omega_j}, \ldots, x_{\omega_j}).$$ The equation ${\scr Q{\bf f}=R}$ expresses all of these constraints in tensor form, so it must hold.
$(\impliedby)$ From above, $f:R \rightarrow R$ defined by $f(c) = \rho(c,\ldots,c)$ has the desired property. This completes the proof.
Discussion
This result allows one to easily reject risk functions which do not have associated latent disutilities, by evaluating $\rho$ for particular $Z$ and enforcing linear constraints. It also provides an explicit construction of the unique $f$ from $\rho$ in cases where such an $f$ exists. This makes it easy to check monotonicity/continuity/convexity.
Unfortunately, even though the condition I present is sufficient, it would be quite difficult to use this result to confirm risk functions which do have associated latent disutilities. I'd love to see if someone who knows more about crazy linear algebra could find a more easily verifiable sufficient condition. And of course, a generalization to infinite sample spaces would be incredible, although at this point I'm afraid I wouldn't understand it without a PhD in real analysis.
Best Answer
Risk aversion is defined as having a lower expected utility from taking a lottery $L=(p_1,x_1;\dots;p_n,x_n)$ than the utility from taking the expected value of that lottery with certainty. Or mathematically, \begin{equation} EU=\sum_{i=1}^np_iu(x_i)<u\left(\sum_{i=1}^np_ix_i\right)=u(EV). \end{equation} This defining condition of risk aversion turns out to be equivalent to requiring that $u(\cdot)$ be strictly concave because of Jensen's inequality.
Unless you want to dispute the definition of risk aversion above, you'd have to live with its implications.
In particular, suppose you have "exponential" love for money, say $u(x)=\mathrm e^x$. Then, we must have \begin{equation} \frac12\mathrm e^1+\frac12\mathrm e^3>\mathrm e^2. \end{equation} Namely, you'd be willing to take a gamble that gives you equal chances of getting an ex post wealth of either $1$ or $3$ over a riskless option that guarantees a wealth level of $2$ (the expected value of the gamble). This behavior can hardly be squared with the usual understanding of risk aversion.