Is there an intuitive argument for the correct bounds of a confidence interval in a one-sided hypothesis test? Assume I have a hypothesis 
$$H_0: \mu \leq 0$$
$$H_1: \mu > 0$$
which corresponds to R's alternativ "greater" and which I would like to check with a confidence level of $95\%$ ($\alpha = 0.05$). I thought the confidence interval (CI) must be  of the form $\left(- \infty, a\right]$, which is wrong. Is there an intuitive argument for the opposite? Long form:
Lets assume we have $n=10$ measuremts with $\bar{x} = -9.603$ and $s_x = 2.342$ as in the R example below. For a two-sided test (without any bias) it is clear that the CI can be calculate to be 
$$\left[\bar{x}-c\cdot\frac{s_x}{\sqrt{10}}, \bar{x}-c\cdot\frac{s_x}{\sqrt{10}}\right] = \left[-11.28, -7.93\right]$$
using the t-distributiuon $F_9(c) = 1 - \frac{\alpha}{2}$.
But in my case I want to consider a one-sided test and I thought about the two options (here $F_9(c) = 1 - \alpha$)


*

*$\left(- \infty, \bar{x}-c\cdot\frac{s_x}{\sqrt{10}}\right]$

*$\left[\bar{x}-c\cdot\frac{s_x}{\sqrt{10}}, +\infty\right)$
At first, my intuition for $H_0: \bar{x} \leq 0$ told me that the first option 
should be correct: "$-\infty$ is less 0". But when looking at the hypothesis test, I realized, I was wrong.
Our test statistics is $t = \frac{\bar{x} - 0}{s_x/\sqrt{n}} = -12.968$ and our cut off value is $c = F^{-1}(1 - \alpha) = 1.833$ resulting in a p-value of $1 - F_9(t) \approx 1$. Thus, we accept the null-hypothesis, because $t < c$ or $p > \alpha$ (as expected from the values!).
If we look at the two intervals, it is clear (the result must be the same!) that the second option is correct. whereas the first is wrong:


*

*$0 \notin \left(-\infty, -8.25\right]$

*$0 \in \left[-10.96, +\infty\right)$

Some references:
(1),
(2) ,
(3) and
(4) (Null hypothesis for directional tests),
Working example in R:
set.seed(1)
conf_level <- 0.95
x <- rnorm(10, -10, 3)
xm <- mean(x)
sx <- sd(x)
print(paste0("Mean = ", xm))
print(paste0("Standard deviation = ", sx))
mu0 <- 0
# H_0: mean(x) < 0 (which is true)
print(t.test(x, mu = 0,alternative = "greater", conf.level = conf_level))

n <- length(x)
t <- (xm - mu0)/(sx/sqrt(n)) # tests statistics
print(paste0("t = ", t))
df <- n - 1
print(paste0("df = ", df))
alpha <- 1 - conf_level # probability type I error

c_val <- qt(1 - alpha, df = df)

p_value <- 1 - pt(t, df)
print(paste0("p_value = ", p_value))

# confidence interval
cint <- qt(1 - alpha, df = df) * sqrt(sx^2/n)
cint <- xm + c(-Inf, cint)
print(paste0("CI1 = (", paste(cint, collapse = ", "), "]"))

cint <- qt(1 - alpha, df = df) * sqrt(sx^2/n)
cint <- xm + c(-cint, Inf)
print(paste0("CI2 = [", paste0(cint, collapse = ", "), ")"))

 A: Suppose you are testing a plan of medical care and diet
that will help badly malnourished babies gain weight.
You have $n = 10$ subjects. After a week on the plan
you measure the change in weight of each child in ounces
(an ounce is about 30g).
Your null hypothesis is that the population mean weight of such
children stays the same or decreases, and the alternative
hypothesis (research hypothesis) is that the mean weight 
increases:  $H_0: \mu \le 0$ vs. $H_a: \mu > 0.$
One difficulty with your question is that you are mixing up
the notation for population parameters with the notation for
sample statistics. Notice that the null and alternative hypotheses
are stated in terms of population parameters.
Upon finding actual weight changes $X_1, X_2, \dots X_{10},$ you find
the following summary statistics for your sample of 10 children.
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
-12.507 -11.639  -9.230  -9.603  -8.339  -5.214 
sd(x)
[1] 2.341758

This is bad news because weight changes are mainly negative. In particular,
the sample mean is $\bar X = \frac 1n \sum_{i=1}^n X_i = -9.603$ and
the sample standard deviation is 
$S = \sqrt{\frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2} = 2.342.$
A 95% (2-sided) confidence interval for the population mean weight gain $\mu$ is of the
form $\bar X \pm t^*\frac{S}{\sqrt{n}},$ where $t^*$ cuts 2.5% of the probability from the upper tail of Student's t distribution with $\nu = n-1$ degrees of freedom. For your data this computes to $(-11.28,-7.93).$
The computation of this confidence interval is included in the t.test procedure in R. (You did the computation correctly, but used the wrong notation.)
t.test(x)$conf.int
[1] -11.278584  -7.928199
 attr(,"conf.level")
 [1] 0.95

If you use R to test $H_0: \mu \le 0$ vs. $H_a: \mu > 0,$ then the syntax and output from R are as shown below:
t.test(x, mu=0, alte="g")  # 'g' for 'greater'

        One Sample t-test

data:  x
t = -12.968, df = 9, p-value = 1
alternative hypothesis: true mean is greater than 0
95 percent confidence interval:
 -10.96086       Inf
sample estimates:
mean of x 
-9.603392 

The P-value is 1. In order to reject $H_0$ at the 5% level (and conclude that the program is
helping the children gain weight), you would need a P-value below 5%.
Ordinarily, a P-value at or near 1 means that something is wrong with the model or the hypothesis. In this case, results are unexpectedly bad and the program is not working as intended. (Unless you have prior experience with such
programs and subjects to warrant an expectation that children will gain weight, it is probably better to start with a two-sided alternative.)
The interpretation of the one-sided 95% confidence interval $(-10.96, \infty)$       in the R output
just above is something like this: "Not only did the the program fail to help the children as intended, the actual change in weight could be a loss
of as much as $10.96$ ounces." 
In this particular example, the data were simulated from the population
$\mathsf{Norm}(\mu = -10, \sigma = 3).$ Usually, using real data, one does
not know the population parameters for sure. We would not know that it was
correct not to reject $H_0;$ we would not know that the 2-sided confidence
interval $(-11.28,-7.93)$ truly does contain $\mu = -10;$ nor would we
know that the one-sided confidence interval $(-10.96, \infty)$ from the one-sided test procedure also contains $\mu = -10.$ 
In summary, there are several issues with your question potentially leading
to confusion:  (1) Incorrect use of $\mu$ and $\sigma$ to represent
sample mean and standard deviation. (2) Unfortunate choice of a one-sided
alternative for the test of hypothesis leading to a huge P-value. (3) The one-sided confidence interval that is included in R output for a right-sided
alternative may not give you the information you most wanted from a confidence interval.
You might find it instructive to go through the steps above starting
with data y = rnorm(10, 10, 3); still using the right-sided alternative
and presumably rejecting $H_0$;  interpreting the accompanying one-sided
confidence interval which now gives a positive lower bound.
It is a nice idea to add print statements to the standard R code that will
include German terminology, but don't use population notation to label sample statistics. Also, as you probably know, most browsers will enable you to
translate this answer to German. I will give that a try in a few minutes to make sure
my English is reasonably compatible with the translation algorithm.
