There are a few slight language mistakes that are creating a bit of confusion regarding your question.
But why should we reason in this way? Each decision, in fact, has a distribution of utility associated with it. Why do we compare the distributions of utilities for different choices only by a single summary statistic? And why do we pick the mean rather than mode or median, etc.?
Utilities do not have a distribution. Outcomes have a distribution and via the outcome, actions, in some cases, have a distribution. Utility is deterministic. If it were random, then your feelings regarding an outcome would constantly startle you. For example, you could have the experience of "wow, having my legs crushed in a motor vehicle accident was a surprisingly good experience!" What is uncertain is the outcome from an action.
If we exclude degenerate cases, where the integrals diverge, and a solution does not exist, I think I can also show you a case where the median maximizes expected utility.
Note that $\mathcal{U}(\delta(x),\mu)=-\mathcal{L}(\delta(x),\mu)$. We find it important to create a rule, which is what we are evaluating with our utility, that will find $\mu$ with some consistency.
We want to solve: $$\min_{\delta}\mathcal{L}(\delta,\mu)=|\delta(x)-\mu|$$ subject to $$f(x|\mu)=\frac{1}{\pi}\frac{1}{1+(x-\mu)^2}.$$
If we assume that $\Pr(\mu)\propto{1},$ then the risk is $$\int_{-\infty}^\infty|\delta(x)-\mu|\prod_{i=1}^n\frac{1}{\pi}\frac{1}{1+(x_i-\mu)^2}\mathrm{d}x$$
and the integrated risk minimizes when $$\int_{-\infty}^\infty\int_{-\infty}^\infty|\delta(x)-\mu|\prod_{i=1}^n\frac{1}{\pi}\frac{1}{1+(x_i-\mu)^2}\mathrm{d}x\mathrm{d}\mu$$ is at a minimum. It minimizes when $\delta(x)$ is the median.
You maximize expected utility when you find the median of the data. You cannot find a mean for $$f(x|\mu)=\frac{1}{\pi}\frac{1}{1+(x-\mu)^2},$$ as it does not exist. Because it has no mean, it also has no variance. Since it has no variance, you cannot minimize quadratic loss. Consequently, quadratic utility, if it were the true case, would be minimized by any value in the real numbers.
If you ignore the degenerate cases, like the above case, expected utility has an unexpected advantage over other methods. Considering all possible decision rules and actions that could be taken, when you use expected utility, then you end up with a total ordering. You are correct, there could be ties, but because the impact of all parameters has been accounted for, you would be indifferent between the choices with tied utility.
The alternative, which is used in Frequentist decision theory, is to order the risk function through stochastic dominance. A Frequentist decision is said to be admissible if it cannot be stochastically dominated. This doesn't permit a total ordering. Nonetheless, if $\delta(x)$ first-order stochastically dominates $\delta'(x)$, then it is also true that the expected utility of choosing $\delta>\delta'$. So the alternative gets you the same result.
There are a few other solutions that can be used, but they either map to maximizing expected utility, or they beg the question of why you would use them in the cases where they do not. To give another statistical example, imagine you read a research study that had a sample size of one million observations using maximum likelihood methods or Bayesian methods. You replicate the study with a sample size of 100 and estimate the mean and the variance using an unbiased estimator. Neither Bayesian nor maximum likelihood estimators are unbiased in the general case.
You insist that you will not combine your estimates because the other estimate is biased, whereas yours in unbiased. Bayesian methods offer a disciplined method to combine the samples into a single point estimator maximizing your utility. You insist on losing the information in the one million person sample in favor of unbiasedness.
Now, if your utility had a very very strong bias toward unbiased estimators, then you would be maximizing your utility by not maximizing the utility of your estimator. But in the absence of that, the biased estimator will be far more accurate than that from your small sample alone. If accuracy maximizes your utility, then you end up choosing an estimator that is utility maximizing.
Don't confuse the expectation of the utility with the expected value of the action. Those are different things.
Also, consider maximizing expected utility versus median utility. You take the utility of every outcome times its probability and sum it. $$\mathbb{E}[\mathcal{U}(\tilde{x})]=\int_{\tilde{x}\in\chi}\mathcal{U}(\tilde{x})\Pr(\tilde{x})\mathrm{d}\tilde{x}$$
Now let's think about the median utility.$$\mathbb{M}[\mathcal{U}(\tilde{x})]=c$$ if $$\int_a^c\mathcal{U}(\tilde{x})\Pr(\tilde{x})\mathrm{d}\tilde{x}=\int_c^b\mathcal{U}(\tilde{x})\Pr(\tilde{x})\mathrm{d}\tilde{x}.
$$
What would that mean? You would be just as happy if you landed to the left as you landed to the right of $c$? Why would you care about that?
If you chose an action that maximized the expected utility, then there is no action that you could take that you believe would make you happier. The median utility doesn't permit a maximization as the action is chosen by force of being in the center. You would always take the action that gives you a fifty percent chance of being happier than usual or sadder than usual. What a strange thing to do!
EDIT
From Kolmogorov's axioms, the sum of a distribution must equal one. Consider a case with two sets of actions, $a$ and $a'$, where $a'$ is the set of actions which are not $a$.
Focusing in on $a$, let us assume that the utility function is $-x^2$. Let us assume that $x$, when the action is $a$, is drawn from $f(x)=\exp(-x),x>0$.
Noting that $$\int_0^\infty\exp(-x)\mathrm{d}x=1,$$ we can readily confirm that it is a probability density function. Including utility results in $$\int_0^\infty{x^2}\exp(-x)\mathrm{d}x=-2,$$ which confirms it is not a distribution. $$\mathbb{E}(\mathcal{U}(a))=-2.$$
While it would be possible to constuct a distribution by utilities, it won't necessarily be a function since if $g(x)=\mathcal{U}(x)\Pr(x)$, then $g^{-1}(x)$ isn't guaranteed to be a function.