How to use WPF DataGrid (.Net 4.0) in PowerShell ISE?

It was a combination of multiple errors.

1. A slash was missing at the end of the tag
2. I had a TYPO inpowershell_ise.exe.Config and the .Net 4.0 assemblies didn't load

3. It's a good to check which assemblies are loaded with

   [System.AppDomain]::CurrentDomain.GetAssemblies() | sort location

Now here is a working solution

function Invoke-sql1
{
    param( [string]$sql,
           [System.Data.SQLClient.SQLConnection]$connection
           )
    $cmd = new-object System.Data.SQLClient.SQLCommand($sql,$connection)
    $ds = New-Object system.Data.DataSet
    $da = New-Object System.Data.SQLClient.SQLDataAdapter($cmd)
    $da.fill($ds) | Out-Null
    return $ds.tables[0].rows
}


[xml] $xaml = @"
<Window 
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        Title="MainWindow" Height="350" Width="525">
    <Grid>
        <DataGrid 
            AutoGenerateColumns="True" 
            HorizontalAlignment="Left" 
            Name="dataGrid1" 
            VerticalAlignment="Top" 
            Width="330"
            HeadersVisibility="All" 
            >
            <DataGrid.Columns>
                <DataGridTextColumn Header="title"
                                Binding="{Binding title}" 
                                />
                <DataGridTextColumn Header="itemid"
                                Binding="{Binding itemid}" 
                                />
            </DataGrid.Columns>
        </DataGrid >
    </Grid>
</Window>
"@
$reader=(New-Object System.Xml.XmlNodeReader $xaml)
$Form=[Windows.Markup.XamlReader]::Load( $reader )


$con = New-Object System.Data.SqlClient.SqlConnection
$con.ConnectionString = "Data Source=localhost;Initial Catalog=ABDATA;Integrated Security=True"
$con.open()

$sql = @"
SELECT 'abc' title, 3 itemid
union
SELECT 'xyz' title, 2 itemid
union
SELECT 'efg' title, 1 itemid
"@

$dg = $Form.FindName("dataGrid1")
$dg.ItemsSource = @(Invoke-sql1 $sql $con)
$Form.ShowDialog() 

The only monor problem is, that I have to define the columns myself. I thought that could be done automatically.